The number of futuristic sci-fi movie in Golden Banana Movie Festival equals 1 plus the number of intense detective thriller in Golden Banana Movie Festival. The number of intense detective thriller in West Sahara Movie Festival equals the number of calm road movie in Golden Banana Movie Festival. The number of futuristic sci-fi movie in Taylor Movie Festival equals 4. The number of calm road movie in Taylor Movie Festival equals the number of intense detective thriller in Taylor Movie Festival. The number of intense detective thriller in Golden Banana Movie Festival equals the total number of movies in Taylor Movie Festival. The number of intense detective thriller in Taylor Movie Festival equals 3 times the number of futuristic sci-fi movie in Taylor Movie Festival. The number of futuristic sci-fi movie in West Sahara Movie Festival equals 3. The number of calm road movie in West Sahara Movie Festival exists, and its number is greater than 0. The number of calm road movie in Golden Banana Movie Festival equals the sum of the number of futuristic sci-fi movie in Golden Banana Movie Festival, the number of calm road movie in Taylor Movie Festival, and the number of futuristic sci-fi movie in Taylor Movie Festival. The total number of movies in West Sahara Movie Festival equals 50.

Question: How many calm road movie does West Sahara Movie Festival have?

Solution: The question is difficult, so we use equations to solve it. Define futuristic sci-fi movie in Taylor Movie Festival as R; so R = 4. Define intense detective thriller in Taylor Movie Festival as g; h = R = 4; so g = 3 * h = (3) * (4) = 12. Define calm road movie in Taylor Movie Festival as W; so W = g = 12. Define total number of movies in Taylor Movie Festival as H; e = g + W = 12 + 12 = 24; so H = e + R = 24 + 4 = 28. Define intense detective thriller in Golden Banana Movie Festival as I; so I = H = 28. Define futuristic sci-fi movie in Golden Banana Movie Festival as m; p = I = 28; so m = 1 + p = 1 + 28 = 29. Define calm road movie in Golden Banana Movie Festival as r; A = m + W = 29 + 12 = 41; so r = A + R = 41 + 4 = 45. Define intense detective thriller in West Sahara Movie Festival as G; so G = r = 45. Define futuristic sci-fi movie in West Sahara Movie Festival as s; so s = 3. Define calm road movie in West Sahara Movie Festival as x; we don't know its value yet but will find it out later. Define total number of movies in West Sahara Movie Festival as k; N = G + x = 45 + x = x + 45; so k = N + s = x + 45 + 3 = x + 48. We know k = 50, so we have x + 48 = 50 and we solve it through the following steps: Simplifying the equation: x + 48 = 50. Move all terms to one side: x - 2 = 0. Isolate the term with x: 1*x = 2. Divide both sides by 1: x = 2 / 1. Solution: x = 2. Answer: 2.


The total number of adult animals in Cedar Valley equals 284. The number of adult fox in Oakridge Riverside equals 4 times the sum of the total number of adult animals in Maple Creek, the number of adult wolf in Maple Creek, and the number of adult fox in Maple Creek. The number of adult fox in Cedar Valley equals the sum of the number of adult wolf in Cedar Valley, the number of adult wolf in Oakridge Riverside, the number of adult wolf in Maple Creek, and the total number of adult animals in Oakridge Riverside. The number of adult wolf in Oakridge Riverside exists, and its number is greater than 0. The number of adult wolf in Maple Creek equals 4. The number of adult wolf in Cedar Valley equals the total number of adult animals in Oakridge Riverside. The number of adult fox in Maple Creek equals 3 plus the number of adult wolf in Maple Creek.

Question: How many adult wolf does Oakridge Riverside have?

Solution: The question is difficult, so we use equations to solve it. Define adult wolf in Maple Creek as N; so N = 4. Define adult fox in Maple Creek as t; e = N = 4; so t = 3 + e = 3 + 4 = 7. Define total number of adult animals in Maple Creek as J; so J = N + t = 4 + 7 = 11. Define adult fox in Oakridge Riverside as j; W = J + N = 11 + 4 = 15; S = W + t = 15 + 7 = 22; so j = 4 * S = (4) * (22) = 88. Define adult wolf in Oakridge Riverside as x; we don't know its value yet but will find it out later. Define total number of adult animals in Oakridge Riverside as m; so m = j + x = 88 + x = x + 88. Define adult wolf in Cedar Valley as C; so C = m = x + 88. Define adult fox in Cedar Valley as G; Z = C + x = x + 88 + x = 2*x + 88; b = Z + N = 2*x + 88 + 4 = 2*x + 92; so G = b + m = 2*x + 92 + x + 88 = 3*x + 180. Define total number of adult animals in Cedar Valley as B; so B = C + G = x + 88 + 3*x + 180 = 4*x + 268. We know B = 284, so we have 4*x + 268 = 284 and we solve it through the following steps: Simplifying the equation: 4*x + 268 = 284. Move all terms to one side: 4*x - 16 = 0. Isolate the term with x: 4*x = 16. Divide both sides by 4: x = 16 / 4. Solution: x = 4. Answer: 4.


The number of intense detective thriller in Verdi Movie Festival exists, and its number is greater than 0. The number of intense detective thriller in Verdi Movie Festival exists, and its number is greater than 0. The number of futuristic sci-fi movie in West Sahara Movie Festival equals 4 times the difference between the number of calm road movie in Golden Banana Movie Festival and the number of intense detective thriller in Verdi Movie Festival. The number of solemn period drama in Taylor Movie Festival equals the number of intense detective thriller in Verdi Movie Festival. The number of intense detective thriller in Verdi Movie Festival exists, and its number is greater than 0. The total number of movies in Northwood Movie Festival equals 148. The number of futuristic sci-fi movie in Northwood Movie Festival equals the number of upbeat metropolis comedy in Verdi Movie Festival. The number of calm road movie in Golden Banana Movie Festival equals 3 plus the number of solemn period drama in Taylor Movie Festival.

Question: How many intense detective thriller does Verdi Movie Festival have?

Solution: The question is difficult, so we use equations to solve it. Define intense detective thriller in Verdi Movie Festival as x; we don't know its value yet but will find it out later. Define solemn period drama in Taylor Movie Festival as f; so f = x = x. Define calm road movie in Golden Banana Movie Festival as g; q = f = x; so g = 3 + q = 3 + x = x + 3. Define futuristic sci-fi movie in West Sahara Movie Festival as y; U = g - x = (x + 3) - (x) = 3; so y = 4 * U = (4) * (3) = 12. Define solemn period drama in Golden Banana Movie Festival as d; L = y + x = 12 + x = x + 12; C = L + f = x + 12 + x = 2*x + 12; so d = 4 + C = 4 + 2*x + 12 = 2*x + 16. Define total number of movies in Golden Banana Movie Festival as r; so r = g + d = x + 3 + 2*x + 16 = 3*x + 19. Define upbeat metropolis comedy in Verdi Movie Festival as V; k = r + x = 3*x + 19 + x = 4*x + 19; e = k + g = 4*x + 19 + x + 3 = 5*x + 22; so V = 4 * e = (4) * (5*x + 22) = 20*x + 88. Define futuristic sci-fi movie in Northwood Movie Festival as o; so o = V = 20*x + 88. Define total number of movies in Northwood Movie Festival as h; so h = o = 20*x + 88. We know h = 148, so we have 20*x + 88 = 148 and we solve it through the following steps: Simplifying the equation: 20*x + 88 = 148. Move all terms to one side: 20*x - 60 = 0. Isolate the term with x: 20*x = 60. Divide both sides by 20: x = 60 / 20. Solution: x = 3. Answer: 3.


The total number of adult animals in Mayer Aquarium equals 188. The number of adult wolf in Bundle Ranch exists, and its number is greater than 0. The number of adult wolf in Jefferson Circus equals 2 times the sum of the total number of adult animals in South Zoo, the number of adult deer in South Zoo, and the number of adult wolf in Bundle Ranch. The number of adult bear in Mayer Aquarium equals 2 times the sum of the total number of adult animals in Jefferson Circus, the number of adult deer in South Zoo, and the number of adult wolf in Bundle Ranch. The number of adult deer in South Zoo equals 4 times the sum of the total number of adult animals in Bundle Ranch and the number of adult wolf in Bundle Ranch. The number of adult racoon in Jefferson Circus equals 4 times the number of adult wolf in Bundle Ranch.

Question: How many adult wolf does Bundle Ranch have?

Solution: The question is difficult, so we use equations to solve it. Define adult wolf in Bundle Ranch as x; we don't know its value yet but will find it out later. Define total number of adult animals in Bundle Ranch as p; so p = x = x. Define adult deer in South Zoo as r; W = p + x = x + x = 2*x; so r = 4 * W = (4) * (2*x) = 8*x. Define total number of adult animals in South Zoo as V; so V = r = 8*x. Define adult wolf in Jefferson Circus as b; H = V + r = 8*x + 8*x = 16*x; d = H + x = 16*x + x = 17*x; so b = 2 * d = (2) * (17*x) = 34*x. Define adult racoon in Jefferson Circus as T; R = x = x; so T = 4 * R = (4) * (x) = 4*x. Define total number of adult animals in Jefferson Circus as y; so y = T + b = 4*x + 34*x = 38*x. Define adult bear in Mayer Aquarium as i; c = y + r = 38*x + 8*x = 46*x; g = c + x = 46*x + x = 47*x; so i = 2 * g = (2) * (47*x) = 94*x. Define total number of adult animals in Mayer Aquarium as S; so S = i = 94*x. We know S = 188, so we have 94*x = 188 and we solve it through the following steps: Simplifying the equation: 94*x = 188. Move all terms to one side: 94*x - 188 = 0. Isolate the term with x: 94*x = 188. Divide both sides by 94: x = 188 / 94. Solution: x = 2. Answer: 2.


The number of calm road movie in Festival Lumière de Valmont exists, and its number is greater than 0. The number of calm road movie in Rêves de Belleville equals the sum of the total number of movies in Cinéma de Montreval, the number of calm road movie in Festival Lumière de Valmont, the total number of movies in Festival Lumière de Valmont, and the number of intense detective thriller in Cinéma de Montreval. The number of intense detective thriller in Rêves de Belleville equals 2 plus the number of solemn period drama in Cinéma de Montreval. The total number of movies in Rêves de Belleville equals 18. The number of intense detective thriller in Cinéma de Montreval equals the number of solemn period drama in Festival Lumière de Valmont. The number of solemn period drama in Festival Lumière de Valmont equals the number of calm road movie in Festival Lumière de Valmont. The number of solemn period drama in Cinéma de Montreval equals 3 plus the total number of movies in Festival Lumière de Valmont. The number of upbeat metropolis comedy in Rêves de Belleville equals the number of solemn period drama in Festival Lumière de Valmont.

Question: How many calm road movie does Festival Lumière de Valmont have?

Solution: The question is difficult, so we use equations to solve it. Define calm road movie in Festival Lumière de Valmont as x; we don't know its value yet but will find it out later. Define solemn period drama in Festival Lumière de Valmont as p; so p = x = x. Define total number of movies in Festival Lumière de Valmont as T; so T = x + p = x + x = 2*x. Define intense detective thriller in Cinéma de Montreval as u; so u = p = x. Define solemn period drama in Cinéma de Montreval as d; N = T = 2*x; so d = 3 + N = 3 + 2*x = 2*x + 3. Define total number of movies in Cinéma de Montreval as U; so U = u + d = x + 2*x + 3 = 3*x + 3. Define calm road movie in Rêves de Belleville as R; Q = U + x = 3*x + 3 + x = 4*x + 3; v = Q + T = 4*x + 3 + 2*x = 6*x + 3; so R = v + u = 6*x + 3 + x = 7*x + 3. Define intense detective thriller in Rêves de Belleville as m; s = d = 2*x + 3; so m = 2 + s = 2 + 2*x + 3 = 2*x + 5. Define upbeat metropolis comedy in Rêves de Belleville as K; so K = p = x. Define total number of movies in Rêves de Belleville as S; g = R + m = 7*x + 3 + 2*x + 5 = 9*x + 8; so S = g + K = 9*x + 8 + x = 10*x + 8. We know S = 18, so we have 10*x + 8 = 18 and we solve it through the following steps: Simplifying the equation: 10*x + 8 = 18. Move all terms to one side: 10*x - 10 = 0. Isolate the term with x: 10*x = 10. Divide both sides by 10: x = 10 / 10. Solution: x = 1. Answer: 1.


The number of adult owl in Maple Creek equals 2 plus the sum of the total number of adult animals in Pine Ridge and the number of adult parrot in Beverly Forest. The number of adult parrot in Pine Ridge equals 4 times the sum of the number of adult parrot in Beverly Forest, the number of adult crow in Pine Ridge, and the number of adult blue jay in Cedar Valley. The total number of adult animals in Maple Creek equals 478. The number of adult blue jay in Cedar Valley exists, and its number is greater than 0. The number of adult parrot in Beverly Forest equals 4 times the sum of the number of adult blue jay in Cedar Valley, the total number of adult animals in Cedar Valley, and the number of adult crow in Pine Ridge. The number of adult crow in Pine Ridge equals 4 plus the sum of the total number of adult animals in Cedar Valley and the number of adult blue jay in Cedar Valley.

Question: How many adult blue jay does Cedar Valley have?

Solution: The question is difficult, so we use equations to solve it. Define adult blue jay in Cedar Valley as x; we don't know its value yet but will find it out later. Define total number of adult animals in Cedar Valley as H; so H = x = x. Define adult crow in Pine Ridge as R; a = H + x = x + x = 2*x; so R = 4 + a = 4 + 2*x = 2*x + 4. Define adult parrot in Beverly Forest as Y; L = x + H = x + x = 2*x; g = L + R = 2*x + 2*x + 4 = 4*x + 4; so Y = 4 * g = (4) * (4*x + 4) = 16*x + 16. Define adult parrot in Pine Ridge as z; f = Y + R = 16*x + 16 + 2*x + 4 = 18*x + 20; u = f + x = 18*x + 20 + x = 19*x + 20; so z = 4 * u = (4) * (19*x + 20) = 76*x + 80. Define total number of adult animals in Pine Ridge as T; so T = R + z = 2*x + 4 + 76*x + 80 = 78*x + 84. Define adult owl in Maple Creek as p; s = T + Y = 78*x + 84 + 16*x + 16 = 94*x + 100; so p = 2 + s = 2 + 94*x + 100 = 94*x + 102. Define total number of adult animals in Maple Creek as j; so j = p = 94*x + 102. We know j = 478, so we have 94*x + 102 = 478 and we solve it through the following steps: Simplifying the equation: 94*x + 102 = 478. Move all terms to one side: 94*x - 376 = 0. Isolate the term with x: 94*x = 376. Divide both sides by 94: x = 376 / 94. Solution: x = 4. Answer: 4.


The total number of adult animals in Maple Creek equals 65. The number of adult blue jay in Beverly Forest equals 1. The number of adult parrot in Beverly Forest equals 3 plus the number of adult blue jay in Beverly Forest. The number of adult owl in Maple Creek equals the sum of the total number of adult animals in Cedar Valley, the number of adult crow in Oakridge Riverside, and the number of adult blue jay in Beverly Forest. The number of adult eagle in Oakridge Riverside exists, and its number is greater than 0. The number of adult crow in Oakridge Riverside equals 2 times the total number of adult animals in Beverly Forest. The number of adult owl in Cedar Valley equals 2 times the sum of the total number of adult animals in Oakridge Riverside, the total number of adult animals in Beverly Forest, and the number of adult crow in Oakridge Riverside.

Question: How many adult eagle does Oakridge Riverside have?

Solution: The question is difficult, so we use equations to solve it. Define adult blue jay in Beverly Forest as P; so P = 1. Define adult parrot in Beverly Forest as N; c = P = 1; so N = 3 + c = 3 + 1 = 4. Define total number of adult animals in Beverly Forest as s; so s = P + N = 1 + 4 = 5. Define adult crow in Oakridge Riverside as o; m = s = 5; so o = 2 * m = (2) * (5) = 10. Define adult eagle in Oakridge Riverside as x; we don't know its value yet but will find it out later. Define total number of adult animals in Oakridge Riverside as C; so C = x + o = x + 10 = x + 10. Define adult owl in Cedar Valley as G; b = C + s = x + 10 + 5 = x + 15; z = b + o = x + 15 + 10 = x + 25; so G = 2 * z = (2) * (x + 25) = 2*x + 50. Define total number of adult animals in Cedar Valley as V; so V = G = 2*x + 50. Define adult owl in Maple Creek as F; H = V + o = 2*x + 50 + 10 = 2*x + 60; so F = H + P = 2*x + 60 + 1 = 2*x + 61. Define total number of adult animals in Maple Creek as t; so t = F = 2*x + 61. We know t = 65, so we have 2*x + 61 = 65 and we solve it through the following steps: Simplifying the equation: 2*x + 61 = 65. Move all terms to one side: 2*x - 4 = 0. Isolate the term with x: 2*x = 4. Divide both sides by 2: x = 4 / 2. Solution: x = 2. Answer: 2.


The number of public highschool in Riverton City equals 1 plus the sum of the total number of schools in Clearwater Bay and the number of culinarian school in Oakbridge City. The number of elementary school in Riverton City equals 4 plus the sum of the total number of schools in Oakbridge City, the number of culinarian school in Oakbridge City, and the number of private middle school in Clearwater Bay. The number of elementary school in Ruby Bay equals the sum of the total number of schools in Oakbridge City and the number of culinarian school in Ruby Bay. The number of regional medical school in Ruby Bay equals the number of culinarian school in Oakbridge City. The total number of schools in Riverton City equals 93. The number of public highschool in Shoreline City exists, and its number is greater than 0. The number of public highschool in Shoreline City exists, and its number is greater than 0. The number of culinarian school in Ruby Bay equals 4. The number of private middle school in Clearwater Bay equals the sum of the total number of schools in Shoreline City and the number of public highschool in Shoreline City.

Question: How many public highschool does Shoreline City have?

Solution: The question is difficult, so we use equations to solve it. Define public highschool in Shoreline City as x; we don't know its value yet but will find it out later. Define total number of schools in Shoreline City as q; so q = x = x. Define private middle school in Clearwater Bay as L; so L = q + x = x + x = 2*x. Define total number of schools in Clearwater Bay as R; so R = L = 2*x. Define culinarian school in Oakbridge City as w; v = R + L = 2*x + 2*x = 4*x; P = v + x = 4*x + x = 5*x; so w = P + q = 5*x + x = 6*x. Define total number of schools in Oakbridge City as S; so S = w = 6*x. Define elementary school in Riverton City as z; p = S + w = 6*x + 6*x = 12*x; Y = p + L = 12*x + 2*x = 14*x; so z = 4 + Y = 4 + 14*x = 14*x + 4. Define public highschool in Riverton City as A; t = R + w = 2*x + 6*x = 8*x; so A = 1 + t = 1 + 8*x = 8*x + 1. Define total number of schools in Riverton City as s; so s = A + z = 8*x + 1 + 14*x + 4 = 22*x + 5. We know s = 93, so we have 22*x + 5 = 93 and we solve it through the following steps: Simplifying the equation: 22*x + 5 = 93. Move all terms to one side: 22*x - 88 = 0. Isolate the term with x: 22*x = 88. Divide both sides by 22: x = 88 / 22. Solution: x = 4. Answer: 4.


The total number of movies in Festival de Clairmont equals 248. The number of solemn period drama in Festival de Saint-Rivage exists, and its number is greater than 0. The number of solemn period drama in Festival de Saint-Rivage exists, and its number is greater than 0. The number of calm road movie in Festival de Clairmont equals 3 times the sum of the number of solemn period drama in Festival de Clairmont and the number of solemn period drama in Rêves de Belleville. The number of solemn period drama in Festival de Saint-Rivage exists, and its number is greater than 0. The number of upbeat metropolis comedy in Festival de Saint-Rivage equals 4 plus the number of solemn period drama in Festival de Saint-Rivage.

Question: How many solemn period drama does Festival de Saint-Rivage have?

Solution: The question is difficult, so we use equations to solve it. Define solemn period drama in Festival de Saint-Rivage as x; we don't know its value yet but will find it out later. Define upbeat metropolis comedy in Festival de Saint-Rivage as j; K = x = x; so j = 4 + K = 4 + x = x + 4. Define total number of movies in Festival de Saint-Rivage as J; so J = j + x = x + 4 + x = 2*x + 4. Define solemn period drama in Rêves de Belleville as d; C = J + x = 2*x + 4 + x = 3*x + 4; Z = C + j = 3*x + 4 + x + 4 = 4*x + 8; so d = 4 + Z = 4 + 4*x + 8 = 4*x + 12. Define total number of movies in Rêves de Belleville as V; so V = d = 4*x + 12. Define solemn period drama in Festival de Clairmont as q; z = V + J = 4*x + 12 + 2*x + 4 = 6*x + 16; N = z + x = 6*x + 16 + x = 7*x + 16; so q = N + j = 7*x + 16 + x + 4 = 8*x + 20. Define calm road movie in Festival de Clairmont as e; v = q + d = 8*x + 20 + 4*x + 12 = 12*x + 32; so e = 3 * v = (3) * (12*x + 32) = 36*x + 96. Define total number of movies in Festival de Clairmont as G; so G = e + q = 36*x + 96 + 8*x + 20 = 44*x + 116. We know G = 248, so we have 44*x + 116 = 248 and we solve it through the following steps: Simplifying the equation: 44*x + 116 = 248. Move all terms to one side: 44*x - 132 = 0. Isolate the term with x: 44*x = 132. Divide both sides by 44: x = 132 / 44. Solution: x = 3. Answer: 3.


The number of elementary school in Hawkesbury equals the sum of the total number of schools in Brightford, the number of elementary school in Brightford, and the number of private middle school in Glenfield City. The total number of schools in Hawkesbury equals 61. The number of elementary school in Glenfield City exists, and its number is greater than 0. The number of elementary school in Glenfield City exists, and its number is greater than 0. The number of elementary school in Glenfield City exists, and its number is greater than 0. The number of private middle school in Glenfield City equals 2 times the number of elementary school in Glenfield City.

Question: How many elementary school does Glenfield City have?

Solution: The question is difficult, so we use equations to solve it. Define elementary school in Glenfield City as x; we don't know its value yet but will find it out later. Define private middle school in Glenfield City as c; v = x = x; so c = 2 * v = (2) * (x) = 2*x. Define total number of schools in Glenfield City as Z; so Z = x + c = x + 2*x = 3*x. Define elementary school in Brightford as b; U = Z + x = 3*x + x = 4*x; l = U + c = 4*x + 2*x = 6*x; so b = 3 + l = 3 + 6*x = 6*x + 3. Define total number of schools in Brightford as A; so A = b = 6*x + 3. Define elementary school in Hawkesbury as S; d = A + b = 6*x + 3 + 6*x + 3 = 12*x + 6; so S = d + c = 12*x + 6 + 2*x = 14*x + 6. Define private middle school in Hawkesbury as t; P = A + c = 6*x + 3 + 2*x = 8*x + 3; h = P + x = 8*x + 3 + x = 9*x + 3; so t = h + Z = 9*x + 3 + 3*x = 12*x + 3. Define total number of schools in Hawkesbury as j; so j = S + t = 14*x + 6 + 12*x + 3 = 26*x + 9. We know j = 61, so we have 26*x + 9 = 61 and we solve it through the following steps: Simplifying the equation: 26*x + 9 = 61. Move all terms to one side: 26*x - 52 = 0. Isolate the term with x: 26*x = 52. Divide both sides by 26: x = 52 / 26. Solution: x = 2. Answer: 2.


The total number of movies in Northwood Movie Festival equals 378. The number of futuristic sci-fi movie in Golden Banana Movie Festival equals 1 plus the number of upbeat metropolis comedy in Golden Banana Movie Festival. The number of upbeat metropolis comedy in Northwood Movie Festival equals 3 times the sum of the total number of movies in Taylor Movie Festival, the number of upbeat metropolis comedy in Taylor Movie Festival, and the number of futuristic sci-fi movie in Golden Banana Movie Festival. The number of futuristic sci-fi movie in Taylor Movie Festival equals 3 times the sum of the number of upbeat metropolis comedy in Golden Banana Movie Festival, the number of futuristic sci-fi movie in Golden Banana Movie Festival, and the number of upbeat metropolis comedy in Taylor Movie Festival. The number of futuristic sci-fi movie in Northwood Movie Festival equals the number of upbeat metropolis comedy in Northwood Movie Festival. The number of upbeat metropolis comedy in Golden Banana Movie Festival exists, and its number is greater than 0. The number of upbeat metropolis comedy in Taylor Movie Festival equals 4 plus the sum of the number of futuristic sci-fi movie in Golden Banana Movie Festival and the number of upbeat metropolis comedy in Golden Banana Movie Festival.

Question: How many upbeat metropolis comedy does Golden Banana Movie Festival have?

Solution: The question is difficult, so we use equations to solve it. Define upbeat metropolis comedy in Golden Banana Movie Festival as x; we don't know its value yet but will find it out later. Define futuristic sci-fi movie in Golden Banana Movie Festival as T; t = x = x; so T = 1 + t = 1 + x = x + 1. Define upbeat metropolis comedy in Taylor Movie Festival as s; N = T + x = x + 1 + x = 2*x + 1; so s = 4 + N = 4 + 2*x + 1 = 2*x + 5. Define futuristic sci-fi movie in Taylor Movie Festival as H; Q = x + T = x + x + 1 = 2*x + 1; n = Q + s = 2*x + 1 + 2*x + 5 = 4*x + 6; so H = 3 * n = (3) * (4*x + 6) = 12*x + 18. Define total number of movies in Taylor Movie Festival as o; so o = s + H = 2*x + 5 + 12*x + 18 = 14*x + 23. Define upbeat metropolis comedy in Northwood Movie Festival as a; j = o + s = 14*x + 23 + 2*x + 5 = 16*x + 28; U = j + T = 16*x + 28 + x + 1 = 17*x + 29; so a = 3 * U = (3) * (17*x + 29) = 51*x + 87. Define futuristic sci-fi movie in Northwood Movie Festival as d; so d = a = 51*x + 87. Define total number of movies in Northwood Movie Festival as w; so w = a + d = 51*x + 87 + 51*x + 87 = 102*x + 174. We know w = 378, so we have 102*x + 174 = 378 and we solve it through the following steps: Simplifying the equation: 102*x + 174 = 378. Move all terms to one side: 102*x - 204 = 0. Isolate the term with x: 102*x = 204. Divide both sides by 102: x = 204 / 102. Solution: x = 2. Answer: 2.


The total number of schools in Riverton City equals 87. The number of elementary school in Shoreline City exists, and its number is greater than 0. The number of public highschool in Shoreline City equals 3. The number of private middle school in Riverton City equals the sum of the total number of schools in Oakbridge City, the number of public highschool in Shoreline City, the number of elementary school in Oakbridge City, and the number of private middle school in Oakbridge City. The number of elementary school in Oakbridge City equals 1 plus the sum of the number of private middle school in Oakbridge City, the number of public highschool in Shoreline City, and the total number of schools in Shoreline City. The number of private middle school in Oakbridge City equals 4 plus the sum of the total number of schools in Shoreline City, the number of public highschool in Shoreline City, and the number of elementary school in Shoreline City.

Question: How many elementary school does Shoreline City have?

Solution: The question is difficult, so we use equations to solve it. Define public highschool in Shoreline City as A; so A = 3. Define elementary school in Shoreline City as x; we don't know its value yet but will find it out later. Define total number of schools in Shoreline City as s; so s = A + x = 3 + x = x + 3. Define private middle school in Oakbridge City as Y; Q = s + A = x + 3 + 3 = x + 6; W = Q + x = x + 6 + x = 2*x + 6; so Y = 4 + W = 4 + 2*x + 6 = 2*x + 10. Define elementary school in Oakbridge City as t; d = Y + A = 2*x + 10 + 3 = 2*x + 13; m = d + s = 2*x + 13 + x + 3 = 3*x + 16; so t = 1 + m = 1 + 3*x + 16 = 3*x + 17. Define total number of schools in Oakbridge City as h; so h = Y + t = 2*x + 10 + 3*x + 17 = 5*x + 27. Define private middle school in Riverton City as w; S = h + A = 5*x + 27 + 3 = 5*x + 30; E = S + t = 5*x + 30 + 3*x + 17 = 8*x + 47; so w = E + Y = 8*x + 47 + 2*x + 10 = 10*x + 57. Define total number of schools in Riverton City as r; so r = w = 10*x + 57. We know r = 87, so we have 10*x + 57 = 87 and we solve it through the following steps: Simplifying the equation: 10*x + 57 = 87. Move all terms to one side: 10*x - 30 = 0. Isolate the term with x: 10*x = 30. Divide both sides by 10: x = 30 / 10. Solution: x = 3. Answer: 3.


The number of intense detective thriller in Verdi Movie Festival equals 2 times the sum of the total number of movies in Golden Banana Movie Festival, the number of calm road movie in Golden Banana Movie Festival, and the number of intense detective thriller in Golden Banana Movie Festival. The number of calm road movie in Golden Banana Movie Festival equals 3 plus the sum of the number of upbeat metropolis comedy in Golden Banana Movie Festival and the number of intense detective thriller in Golden Banana Movie Festival. The number of upbeat metropolis comedy in Golden Banana Movie Festival equals 4 plus the number of intense detective thriller in Golden Banana Movie Festival. The number of upbeat metropolis comedy in West Sahara Movie Festival equals 2 plus the sum of the total number of movies in Verdi Movie Festival, the number of intense detective thriller in Verdi Movie Festival, and the number of upbeat metropolis comedy in Golden Banana Movie Festival. The number of intense detective thriller in Golden Banana Movie Festival exists, and its number is greater than 0. The total number of movies in West Sahara Movie Festival equals 194.

Question: How many intense detective thriller does Golden Banana Movie Festival have?

Solution: The question is difficult, so we use equations to solve it. Define intense detective thriller in Golden Banana Movie Festival as x; we don't know its value yet but will find it out later. Define upbeat metropolis comedy in Golden Banana Movie Festival as k; q = x = x; so k = 4 + q = 4 + x = x + 4. Define calm road movie in Golden Banana Movie Festival as Q; c = k + x = x + 4 + x = 2*x + 4; so Q = 3 + c = 3 + 2*x + 4 = 2*x + 7. Define total number of movies in Golden Banana Movie Festival as d; A = k + x = x + 4 + x = 2*x + 4; so d = A + Q = 2*x + 4 + 2*x + 7 = 4*x + 11. Define intense detective thriller in Verdi Movie Festival as E; p = d + Q = 4*x + 11 + 2*x + 7 = 6*x + 18; H = p + x = 6*x + 18 + x = 7*x + 18; so E = 2 * H = (2) * (7*x + 18) = 14*x + 36. Define total number of movies in Verdi Movie Festival as l; so l = E = 14*x + 36. Define upbeat metropolis comedy in West Sahara Movie Festival as r; F = l + E = 14*x + 36 + 14*x + 36 = 28*x + 72; g = F + k = 28*x + 72 + x + 4 = 29*x + 76; so r = 2 + g = 2 + 29*x + 76 = 29*x + 78. Define total number of movies in West Sahara Movie Festival as z; so z = r = 29*x + 78. We know z = 194, so we have 29*x + 78 = 194 and we solve it through the following steps: Simplifying the equation: 29*x + 78 = 194. Move all terms to one side: 29*x - 116 = 0. Isolate the term with x: 29*x = 116. Divide both sides by 29: x = 116 / 29. Solution: x = 4. Answer: 4.


The number of solemn period drama in Festival de Saint-Rivage exists, and its number is greater than 0. The number of solemn period drama in Festival de Saint-Rivage exists, and its number is greater than 0. The number of intense detective thriller in Festival Lumière de Valmont equals 2 times the sum of the total number of movies in Rêves de Belleville and the number of futuristic sci-fi movie in Festival de Saint-Rivage. The total number of movies in Festival Lumière de Valmont equals 140. The number of upbeat metropolis comedy in Rêves de Belleville equals 4 plus the sum of the total number of movies in Festival de Clairmont, the number of upbeat metropolis comedy in Festival de Clairmont, and the number of solemn period drama in Festival de Saint-Rivage. The number of futuristic sci-fi movie in Festival de Saint-Rivage equals 2 times the number of solemn period drama in Festival de Saint-Rivage.

Question: How many solemn period drama does Festival de Saint-Rivage have?

Solution: The question is difficult, so we use equations to solve it. Define solemn period drama in Festival de Saint-Rivage as x; we don't know its value yet but will find it out later. Define futuristic sci-fi movie in Festival de Saint-Rivage as j; d = x = x; so j = 2 * d = (2) * (x) = 2*x. Define total number of movies in Festival de Saint-Rivage as a; so a = x + j = x + 2*x = 3*x. Define upbeat metropolis comedy in Festival de Clairmont as S; f = a + x = 3*x + x = 4*x; M = f + j = 4*x + 2*x = 6*x; so S = 3 + M = 3 + 6*x = 6*x + 3. Define total number of movies in Festival de Clairmont as C; so C = S = 6*x + 3. Define upbeat metropolis comedy in Rêves de Belleville as g; Y = C + S = 6*x + 3 + 6*x + 3 = 12*x + 6; U = Y + x = 12*x + 6 + x = 13*x + 6; so g = 4 + U = 4 + 13*x + 6 = 13*x + 10. Define total number of movies in Rêves de Belleville as q; so q = g = 13*x + 10. Define intense detective thriller in Festival Lumière de Valmont as k; m = q + j = 13*x + 10 + 2*x = 15*x + 10; so k = 2 * m = (2) * (15*x + 10) = 30*x + 20. Define total number of movies in Festival Lumière de Valmont as F; so F = k = 30*x + 20. We know F = 140, so we have 30*x + 20 = 140 and we solve it through the following steps: Simplifying the equation: 30*x + 20 = 140. Move all terms to one side: 30*x - 120 = 0. Isolate the term with x: 30*x = 120. Divide both sides by 30: x = 120 / 30. Solution: x = 4. Answer: 4.


The number of adult owl in Pine Ridge exists, and its number is greater than 0. The number of adult owl in Pine Ridge exists, and its number is greater than 0. The number of adult owl in Maple Creek equals 3 times the number of adult crow in Maple Creek. The number of adult crow in Maple Creek equals 3 plus the total number of adult animals in Cedar Valley. The number of adult parrot in Maple Creek equals the number of adult crow in Maple Creek. The total number of adult animals in Maple Creek equals 455. The number of adult crow in Cedar Valley equals 3 times the sum of the total number of adult animals in Pine Ridge and the number of adult owl in Pine Ridge.

Question: How many adult owl does Pine Ridge have?

Solution: The question is difficult, so we use equations to solve it. Define adult owl in Pine Ridge as x; we don't know its value yet but will find it out later. Define total number of adult animals in Pine Ridge as D; so D = x = x. Define adult crow in Cedar Valley as h; b = D + x = x + x = 2*x; so h = 3 * b = (3) * (2*x) = 6*x. Define adult parrot in Cedar Valley as A; R = h + x = 6*x + x = 7*x; G = R + D = 7*x + x = 8*x; so A = 2 * G = (2) * (8*x) = 16*x. Define total number of adult animals in Cedar Valley as k; so k = A + h = 16*x + 6*x = 22*x. Define adult crow in Maple Creek as K; J = k = 22*x; so K = 3 + J = 3 + 22*x = 22*x + 3. Define adult parrot in Maple Creek as M; so M = K = 22*x + 3. Define adult owl in Maple Creek as i; s = K = 22*x + 3; so i = 3 * s = (3) * (22*x + 3) = 66*x + 9. Define total number of adult animals in Maple Creek as Z; u = M + i = 22*x + 3 + 66*x + 9 = 88*x + 12; so Z = u + K = 88*x + 12 + 22*x + 3 = 110*x + 15. We know Z = 455, so we have 110*x + 15 = 455 and we solve it through the following steps: Simplifying the equation: 110*x + 15 = 455. Move all terms to one side: 110*x - 440 = 0. Isolate the term with x: 110*x = 440. Divide both sides by 110: x = 440 / 110. Solution: x = 4. Answer: 4.


The number of upbeat metropolis comedy in Festival de Clairmont equals 1 plus the number of futuristic sci-fi movie in Festival de Clairmont. The number of futuristic sci-fi movie in Festival de Clairmont exists, and its number is greater than 0. The total number of movies in Rêves de Belleville equals 31. The number of futuristic sci-fi movie in Festival de Clairmont exists, and its number is greater than 0. The number of futuristic sci-fi movie in Festival de Clairmont exists, and its number is greater than 0. The number of upbeat metropolis comedy in Rêves de Belleville equals 3 plus the sum of the number of upbeat metropolis comedy in Festival de Clairmont and the total number of movies in Festival Lumière de Valmont.

Question: How many futuristic sci-fi movie does Festival de Clairmont have?

Solution: The question is difficult, so we use equations to solve it. Define futuristic sci-fi movie in Festival de Clairmont as x; we don't know its value yet but will find it out later. Define upbeat metropolis comedy in Festival de Clairmont as r; M = x = x; so r = 1 + M = 1 + x = x + 1. Define total number of movies in Festival de Clairmont as u; so u = r + x = x + 1 + x = 2*x + 1. Define upbeat metropolis comedy in Festival Lumière de Valmont as a; S = u + x = 2*x + 1 + x = 3*x + 1; j = S + r = 3*x + 1 + x + 1 = 4*x + 2; so a = 4 + j = 4 + 4*x + 2 = 4*x + 6. Define total number of movies in Festival Lumière de Valmont as o; so o = a = 4*x + 6. Define intense detective thriller in Rêves de Belleville as Z; z = o + u = 4*x + 6 + 2*x + 1 = 6*x + 7; m = z + x = 6*x + 7 + x = 7*x + 7; so Z = m + r = 7*x + 7 + x + 1 = 8*x + 8. Define upbeat metropolis comedy in Rêves de Belleville as W; J = r + o = x + 1 + 4*x + 6 = 5*x + 7; so W = 3 + J = 3 + 5*x + 7 = 5*x + 10. Define total number of movies in Rêves de Belleville as w; so w = Z + W = 8*x + 8 + 5*x + 10 = 13*x + 18. We know w = 31, so we have 13*x + 18 = 31 and we solve it through the following steps: Simplifying the equation: 13*x + 18 = 31. Move all terms to one side: 13*x - 13 = 0. Isolate the term with x: 13*x = 13. Divide both sides by 13: x = 13 / 13. Solution: x = 1. Answer: 1.


The total number of adult animals in Jefferson Circus equals 70. The number of adult wolf in Bundle Ranch equals the sum of the number of adult deer in Bundle Ranch, the total number of adult animals in Hamilton Farm, and the number of adult deer in Hamilton Farm. The number of adult deer in Bundle Ranch equals 4 plus the sum of the total number of adult animals in Hamilton Farm and the number of adult deer in Hamilton Farm. The number of adult deer in Hamilton Farm exists, and its number is greater than 0. The number of adult deer in Hamilton Farm exists, and its number is greater than 0. The number of adult fox in Mayer Aquarium equals the sum of the total number of adult animals in Bundle Ranch, the number of adult deer in Bundle Ranch, and the number of adult wolf in Bundle Ranch.

Question: How many adult deer does Hamilton Farm have?

Solution: The question is difficult, so we use equations to solve it. Define adult deer in Hamilton Farm as x; we don't know its value yet but will find it out later. Define total number of adult animals in Hamilton Farm as A; so A = x = x. Define adult deer in Bundle Ranch as V; S = A + x = x + x = 2*x; so V = 4 + S = 4 + 2*x = 2*x + 4. Define adult wolf in Bundle Ranch as r; T = V + A = 2*x + 4 + x = 3*x + 4; so r = T + x = 3*x + 4 + x = 4*x + 4. Define total number of adult animals in Bundle Ranch as k; so k = V + r = 2*x + 4 + 4*x + 4 = 6*x + 8. Define adult fox in Mayer Aquarium as N; q = k + V = 6*x + 8 + 2*x + 4 = 8*x + 12; so N = q + r = 8*x + 12 + 4*x + 4 = 12*x + 16. Define total number of adult animals in Mayer Aquarium as Q; so Q = N = 12*x + 16. Define adult bear in Jefferson Circus as M; j = Q + V = 12*x + 16 + 2*x + 4 = 14*x + 20; J = j + x = 14*x + 20 + x = 15*x + 20; so M = J + k = 15*x + 20 + 6*x + 8 = 21*x + 28. Define total number of adult animals in Jefferson Circus as i; so i = M = 21*x + 28. We know i = 70, so we have 21*x + 28 = 70 and we solve it through the following steps: Simplifying the equation: 21*x + 28 = 70. Move all terms to one side: 21*x - 42 = 0. Isolate the term with x: 21*x = 42. Divide both sides by 21: x = 42 / 21. Solution: x = 2. Answer: 2.


The number of adult eagle in Oakridge Riverside equals 2 plus the number of adult owl in Oakridge Riverside. The number of adult crow in Pine Ridge equals the sum of the total number of adult animals in Maple Creek, the number of adult owl in Oakridge Riverside, the number of adult eagle in Oakridge Riverside, and the total number of adult animals in Oakridge Riverside. The total number of adult animals in Pine Ridge equals 96. The number of adult eagle in Pine Ridge equals the sum of the number of adult crow in Pine Ridge, the number of adult eagle in Oakridge Riverside, and the number of adult owl in Oakridge Riverside. The number of adult owl in Oakridge Riverside exists, and its number is greater than 0. The number of adult owl in Oakridge Riverside exists, and its number is greater than 0.

Question: How many adult owl does Oakridge Riverside have?

Solution: The question is difficult, so we use equations to solve it. Define adult owl in Oakridge Riverside as x; we don't know its value yet but will find it out later. Define adult eagle in Oakridge Riverside as v; s = x = x; so v = 2 + s = 2 + x = x + 2. Define total number of adult animals in Oakridge Riverside as r; so r = x + v = x + x + 2 = 2*x + 2. Define adult blue jay in Maple Creek as E; q = r + x = 2*x + 2 + x = 3*x + 2; W = q + v = 3*x + 2 + x + 2 = 4*x + 4; so E = 3 + W = 3 + 4*x + 4 = 4*x + 7. Define total number of adult animals in Maple Creek as j; so j = E = 4*x + 7. Define adult crow in Pine Ridge as e; B = j + x = 4*x + 7 + x = 5*x + 7; z = B + v = 5*x + 7 + x + 2 = 6*x + 9; so e = z + r = 6*x + 9 + 2*x + 2 = 8*x + 11. Define adult eagle in Pine Ridge as f; N = e + v = 8*x + 11 + x + 2 = 9*x + 13; so f = N + x = 9*x + 13 + x = 10*x + 13. Define total number of adult animals in Pine Ridge as Z; so Z = f + e = 10*x + 13 + 8*x + 11 = 18*x + 24. We know Z = 96, so we have 18*x + 24 = 96 and we solve it through the following steps: Simplifying the equation: 18*x + 24 = 96. Move all terms to one side: 18*x - 72 = 0. Isolate the term with x: 18*x = 72. Divide both sides by 18: x = 72 / 18. Solution: x = 4. Answer: 4.


The number of upbeat metropolis comedy in Festival de Saint-Rivage equals 1 plus the difference between the total number of movies in Rêves de Belleville and the number of upbeat metropolis comedy in Rêves de Belleville. The number of calm road movie in Rêves de Belleville equals 1 plus the number of upbeat metropolis comedy in Rêves de Belleville. The number of upbeat metropolis comedy in Rêves de Belleville equals 2. The number of intense detective thriller in Rêves de Belleville equals the difference between the number of calm road movie in Rêves de Belleville and the number of upbeat metropolis comedy in Rêves de Belleville. The number of calm road movie in Festival de Saint-Rivage equals the sum of the number of upbeat metropolis comedy in Rêves de Belleville and the number of intense detective thriller in Rêves de Belleville. The number of intense detective thriller in Festival de Saint-Rivage equals 4 times the difference between the number of upbeat metropolis comedy in Festival de Saint-Rivage and the number of intense detective thriller in Rêves de Belleville.

Question: What is the total number of movies in Festival de Saint-Rivage?

Solution: Define upbeat metropolis comedy in Rêves de Belleville as T; so T = 2. Define calm road movie in Rêves de Belleville as m; e = T = 2; so m = 1 + e = 1 + 2 = 3. Define intense detective thriller in Rêves de Belleville as L; so L = m - T = 3 - 2 = 1. Define total number of movies in Rêves de Belleville as X; r = T + L = 2 + 1 = 3; so X = r + m = 3 + 3 = 6. Define upbeat metropolis comedy in Festival de Saint-Rivage as Q; I = X - T = 6 - 2 = 4; so Q = 1 + I = 1 + 4 = 5. Define calm road movie in Festival de Saint-Rivage as J; so J = T + L = 2 + 1 = 3. Define intense detective thriller in Festival de Saint-Rivage as H; g = Q - L = 5 - 1 = 4; so H = 4 * g = 4 * 4 = 16. Define total number of movies in Festival de Saint-Rivage as E; D = J + Q = 3 + 5 = 8; so E = D + H = 8 + 16 = 24. Answer: 24.


The number of intense detective thriller in West Sahara Movie Festival equals 2 plus the sum of the total number of movies in Golden Banana Movie Festival, the total number of movies in Verdi Movie Festival, and the number of futuristic sci-fi movie in Golden Banana Movie Festival. The number of calm road movie in Golden Banana Movie Festival equals 4 times the total number of movies in Verdi Movie Festival. The total number of movies in West Sahara Movie Festival equals 190. The number of upbeat metropolis comedy in Verdi Movie Festival equals 3 times the sum of the total number of movies in Northwood Movie Festival and the number of futuristic sci-fi movie in Northwood Movie Festival. The number of futuristic sci-fi movie in Northwood Movie Festival exists, and its number is greater than 0. The number of futuristic sci-fi movie in Golden Banana Movie Festival equals the sum of the number of futuristic sci-fi movie in Northwood Movie Festival, the total number of movies in Northwood Movie Festival, the number of calm road movie in Golden Banana Movie Festival, and the number of upbeat metropolis comedy in Verdi Movie Festival.

Question: How many futuristic sci-fi movie does Northwood Movie Festival have?

Solution: The question is difficult, so we use equations to solve it. Define futuristic sci-fi movie in Northwood Movie Festival as x; we don't know its value yet but will find it out later. Define total number of movies in Northwood Movie Festival as m; so m = x = x. Define upbeat metropolis comedy in Verdi Movie Festival as L; g = m + x = x + x = 2*x; so L = 3 * g = (3) * (2*x) = 6*x. Define total number of movies in Verdi Movie Festival as C; so C = L = 6*x. Define calm road movie in Golden Banana Movie Festival as F; K = C = 6*x; so F = 4 * K = (4) * (6*x) = 24*x. Define futuristic sci-fi movie in Golden Banana Movie Festival as u; B = x + m = x + x = 2*x; R = B + F = 2*x + 24*x = 26*x; so u = R + L = 26*x + 6*x = 32*x. Define total number of movies in Golden Banana Movie Festival as p; so p = F + u = 24*x + 32*x = 56*x. Define intense detective thriller in West Sahara Movie Festival as q; a = p + C = 56*x + 6*x = 62*x; Q = a + u = 62*x + 32*x = 94*x; so q = 2 + Q = 2 + 94*x = 94*x + 2. Define total number of movies in West Sahara Movie Festival as t; so t = q = 94*x + 2. We know t = 190, so we have 94*x + 2 = 190 and we solve it through the following steps: Simplifying the equation: 94*x + 2 = 190. Move all terms to one side: 94*x - 188 = 0. Isolate the term with x: 94*x = 188. Divide both sides by 94: x = 188 / 94. Solution: x = 2. Answer: 2.


The number of adult fox in Bundle Ranch equals 4 plus the difference between the total number of adult animals in South Zoo and the number of adult fox in South Zoo. The number of adult bear in Jefferson Circus equals 1 plus the sum of the number of adult deer in Mayer Aquarium and the number of adult fox in Bundle Ranch. The total number of adult animals in Jefferson Circus equals 61. The number of adult fox in South Zoo exists, and its number is greater than 0. The number of adult fox in South Zoo exists, and its number is greater than 0. The number of adult wolf in Jefferson Circus equals the sum of the total number of adult animals in Mayer Aquarium, the number of adult fox in Bundle Ranch, the total number of adult animals in South Zoo, and the number of adult deer in Mayer Aquarium.

Question: How many adult fox does South Zoo have?

Solution: The question is difficult, so we use equations to solve it. Define adult fox in South Zoo as x; we don't know its value yet but will find it out later. Define total number of adult animals in South Zoo as U; so U = x = x. Define adult fox in Bundle Ranch as S; p = U - x = (x) - (x) = 0; so S = 4 + p = 4 + 0 = 4. Define adult deer in Mayer Aquarium as q; d = S + x = 4 + x = x + 4; O = d + U = x + 4 + x = 2*x + 4; so q = 4 + O = 4 + 2*x + 4 = 2*x + 8. Define total number of adult animals in Mayer Aquarium as B; so B = q = 2*x + 8. Define adult wolf in Jefferson Circus as f; m = B + S = 2*x + 8 + 4 = 2*x + 12; a = m + U = 2*x + 12 + x = 3*x + 12; so f = a + q = 3*x + 12 + 2*x + 8 = 5*x + 20. Define adult bear in Jefferson Circus as V; J = q + S = 2*x + 8 + 4 = 2*x + 12; so V = 1 + J = 1 + 2*x + 12 = 2*x + 13. Define total number of adult animals in Jefferson Circus as G; so G = f + V = 5*x + 20 + 2*x + 13 = 7*x + 33. We know G = 61, so we have 7*x + 33 = 61 and we solve it through the following steps: Simplifying the equation: 7*x + 33 = 61. Move all terms to one side: 7*x - 28 = 0. Isolate the term with x: 7*x = 28. Divide both sides by 7: x = 28 / 7. Solution: x = 4. Answer: 4.


The number of private middle school in Brightford equals 1 plus the number of elementary school in Brightford. The total number of schools in Hawkesbury equals 67. The number of public highschool in Brightford exists, and its number is greater than 0. The number of public highschool in Brightford exists, and its number is greater than 0. The number of private middle school in Hawkesbury equals the sum of the total number of schools in Brightford, the number of elementary school in Brightford, and the number of public highschool in Brightford. The number of public highschool in Hawkesbury equals the sum of the number of private middle school in Hawkesbury and the total number of schools in Brightford. The number of elementary school in Brightford equals 4.

Question: How many public highschool does Brightford have?

Solution: The question is difficult, so we use equations to solve it. Define elementary school in Brightford as U; so U = 4. Define private middle school in Brightford as k; f = U = 4; so k = 1 + f = 1 + 4 = 5. Define public highschool in Brightford as x; we don't know its value yet but will find it out later. Define total number of schools in Brightford as O; j = x + U = x + 4 = x + 4; so O = j + k = x + 4 + 5 = x + 9. Define private middle school in Hawkesbury as I; c = O + U = x + 9 + 4 = x + 13; so I = c + x = x + 13 + x = 2*x + 13. Define elementary school in Hawkesbury as s; B = O + k = x + 9 + 5 = x + 14; l = B + x = x + 14 + x = 2*x + 14; so s = l + U = 2*x + 14 + 4 = 2*x + 18. Define public highschool in Hawkesbury as N; so N = I + O = 2*x + 13 + x + 9 = 3*x + 22. Define total number of schools in Hawkesbury as h; C = s + I = 2*x + 18 + 2*x + 13 = 4*x + 31; so h = C + N = 4*x + 31 + 3*x + 22 = 7*x + 53. We know h = 67, so we have 7*x + 53 = 67 and we solve it through the following steps: Simplifying the equation: 7*x + 53 = 67. Move all terms to one side: 7*x - 14 = 0. Isolate the term with x: 7*x = 14. Divide both sides by 7: x = 14 / 7. Solution: x = 2. Answer: 2.


The number of calm road movie in Rêves de Belleville equals 4 times the difference between the total number of movies in Cinéma de Montreval and the number of calm road movie in Cinéma de Montreval. The number of calm road movie in Cinéma de Montreval equals 3 times the number of solemn period drama in Cinéma de Montreval. The number of intense detective thriller in Festival de Clairmont equals the sum of the total number of movies in Rêves de Belleville, the number of calm road movie in Rêves de Belleville, the number of solemn period drama in Festival de Clairmont, and the number of calm road movie in Cinéma de Montreval. The number of solemn period drama in Cinéma de Montreval exists, and its number is greater than 0. The total number of movies in Festival de Clairmont equals 82. The number of solemn period drama in Festival de Clairmont equals the sum of the total number of movies in Rêves de Belleville, the total number of movies in Cinéma de Montreval, the number of calm road movie in Rêves de Belleville, and the number of calm road movie in Cinéma de Montreval.

Question: How many solemn period drama does Cinéma de Montreval have?

Solution: The question is difficult, so we use equations to solve it. Define solemn period drama in Cinéma de Montreval as x; we don't know its value yet but will find it out later. Define calm road movie in Cinéma de Montreval as t; E = x = x; so t = 3 * E = (3) * (x) = 3*x. Define total number of movies in Cinéma de Montreval as V; so V = x + t = x + 3*x = 4*x. Define calm road movie in Rêves de Belleville as L; c = V - t = (4*x) - (3*x) = x; so L = 4 * c = (4) * (x) = 4*x. Define total number of movies in Rêves de Belleville as l; so l = L = 4*x. Define solemn period drama in Festival de Clairmont as B; a = l + V = 4*x + 4*x = 8*x; H = a + L = 8*x + 4*x = 12*x; so B = H + t = 12*x + 3*x = 15*x. Define intense detective thriller in Festival de Clairmont as K; f = l + L = 4*x + 4*x = 8*x; d = f + B = 8*x + 15*x = 23*x; so K = d + t = 23*x + 3*x = 26*x. Define total number of movies in Festival de Clairmont as C; so C = B + K = 15*x + 26*x = 41*x. We know C = 82, so we have 41*x = 82 and we solve it through the following steps: Simplifying the equation: 41*x = 82. Move all terms to one side: 41*x - 82 = 0. Isolate the term with x: 41*x = 82. Divide both sides by 41: x = 82 / 41. Solution: x = 2. Answer: 2.


The number of public highschool in Oakbridge City equals 4 plus the sum of the total number of schools in Ruby Bay and the number of public highschool in Ruby Bay. The number of public highschool in Ruby Bay exists, and its number is greater than 0. The number of public highschool in Shoreline City equals the sum of the number of culinarian school in Shoreline City, the total number of schools in Ruby Bay, the number of public highschool in Oakbridge City, and the total number of schools in Oakbridge City. The total number of schools in Shoreline City equals 476. The number of regional medical school in Shoreline City equals 4 times the number of public highschool in Shoreline City. The number of culinarian school in Shoreline City equals 2 times the sum of the total number of schools in Oakbridge City, the number of public highschool in Oakbridge City, and the total number of schools in Ruby Bay.

Question: How many public highschool does Ruby Bay have?

Solution: The question is difficult, so we use equations to solve it. Define public highschool in Ruby Bay as x; we don't know its value yet but will find it out later. Define total number of schools in Ruby Bay as s; so s = x = x. Define public highschool in Oakbridge City as D; j = s + x = x + x = 2*x; so D = 4 + j = 4 + 2*x = 2*x + 4. Define total number of schools in Oakbridge City as g; so g = D = 2*x + 4. Define culinarian school in Shoreline City as d; k = g + D = 2*x + 4 + 2*x + 4 = 4*x + 8; H = k + s = 4*x + 8 + x = 5*x + 8; so d = 2 * H = (2) * (5*x + 8) = 10*x + 16. Define public highschool in Shoreline City as m; a = d + s = 10*x + 16 + x = 11*x + 16; i = a + D = 11*x + 16 + 2*x + 4 = 13*x + 20; so m = i + g = 13*x + 20 + 2*x + 4 = 15*x + 24. Define regional medical school in Shoreline City as l; G = m = 15*x + 24; so l = 4 * G = (4) * (15*x + 24) = 60*x + 96. Define total number of schools in Shoreline City as q; W = m + d = 15*x + 24 + 10*x + 16 = 25*x + 40; so q = W + l = 25*x + 40 + 60*x + 96 = 85*x + 136. We know q = 476, so we have 85*x + 136 = 476 and we solve it through the following steps: Simplifying the equation: 85*x + 136 = 476. Move all terms to one side: 85*x - 340 = 0. Isolate the term with x: 85*x = 340. Divide both sides by 85: x = 340 / 85. Solution: x = 4. Answer: 4.


The total number of adult animals in Jefferson Circus equals 72. The number of adult wolf in Bundle Ranch equals 3. The number of adult deer in South Zoo exists, and its number is greater than 0. The number of adult deer in Bundle Ranch equals the number of adult deer in Jefferson Circus. The number of adult deer in Jefferson Circus equals 3. The number of adult wolf in Jefferson Circus equals the sum of the total number of adult animals in South Zoo, the total number of adult animals in Bundle Ranch, and the number of adult fox in South Zoo. The number of adult wolf in South Zoo equals the sum of the number of adult deer in Jefferson Circus, the number of adult fox in South Zoo, and the number of adult deer in Bundle Ranch. The number of adult fox in South Zoo equals 3 times the total number of adult animals in Bundle Ranch.

Question: How many adult deer does South Zoo have?

Solution: The question is difficult, so we use equations to solve it. Define adult deer in Jefferson Circus as v; so v = 3. Define adult wolf in Bundle Ranch as w; so w = 3. Define adult deer in Bundle Ranch as D; so D = v = 3. Define total number of adult animals in Bundle Ranch as b; so b = w + D = 3 + 3 = 6. Define adult fox in South Zoo as o; Q = b = 6; so o = 3 * Q = (3) * (6) = 18. Define adult deer in South Zoo as x; we don't know its value yet but will find it out later. Define adult wolf in South Zoo as T; M = v + o = 3 + 18 = 21; so T = M + D = 21 + 3 = 24. Define total number of adult animals in South Zoo as n; V = o + T = 18 + 24 = 42; so n = V + x = 42 + x = x + 42. Define adult wolf in Jefferson Circus as k; F = n + b = x + 42 + 6 = x + 48; so k = F + o = x + 48 + 18 = x + 66. Define total number of adult animals in Jefferson Circus as z; so z = k + v = x + 66 + 3 = x + 69. We know z = 72, so we have x + 69 = 72 and we solve it through the following steps: Simplifying the equation: x + 69 = 72. Move all terms to one side: x - 3 = 0. Isolate the term with x: 1*x = 3. Divide both sides by 1: x = 3 / 1. Solution: x = 3. Answer: 3.


The number of adult racoon in Oakridge Riverside equals 4. The number of adult racoon in Beverly Forest equals 4 times the number of adult racoon in Cedar Valley. The number of adult wolf in Cedar Valley equals 2 plus the total number of adult animals in Oakridge Riverside. The number of adult racoon in Cedar Valley exists, and its number is greater than 0. The number of adult wolf in Beverly Forest equals the sum of the total number of adult animals in Cedar Valley, the number of adult fox in Oakridge Riverside, the total number of adult animals in Oakridge Riverside, and the number of adult racoon in Beverly Forest. The total number of adult animals in Beverly Forest equals 58. The number of adult fox in Oakridge Riverside equals 3. The number of adult fox in Beverly Forest equals the number of adult racoon in Beverly Forest.

Question: How many adult racoon does Cedar Valley have?

Solution: The question is difficult, so we use equations to solve it. Define adult fox in Oakridge Riverside as U; so U = 3. Define adult racoon in Oakridge Riverside as Z; so Z = 4. Define total number of adult animals in Oakridge Riverside as R; so R = Z + U = 4 + 3 = 7. Define adult wolf in Cedar Valley as n; v = R = 7; so n = 2 + v = 2 + 7 = 9. Define adult racoon in Cedar Valley as x; we don't know its value yet but will find it out later. Define adult racoon in Beverly Forest as A; H = x = x; so A = 4 * H = (4) * (x) = 4*x. Define total number of adult animals in Cedar Valley as O; so O = n + x = 9 + x = x + 9. Define adult wolf in Beverly Forest as W; D = O + U = x + 9 + 3 = x + 12; L = D + R = x + 12 + 7 = x + 19; so W = L + A = x + 19 + 4*x = 5*x + 19. Define adult fox in Beverly Forest as s; so s = A = 4*x. Define total number of adult animals in Beverly Forest as f; q = W + A = 5*x + 19 + 4*x = 9*x + 19; so f = q + s = 9*x + 19 + 4*x = 13*x + 19. We know f = 58, so we have 13*x + 19 = 58 and we solve it through the following steps: Simplifying the equation: 13*x + 19 = 58. Move all terms to one side: 13*x - 39 = 0. Isolate the term with x: 13*x = 39. Divide both sides by 13: x = 39 / 13. Solution: x = 3. Answer: 3.


The number of private middle school in Riverton City equals 1 plus the number of elementary school in Riverton City. The number of private middle school in Shoreline City equals the number of regional medical school in Riverton City. The number of regional medical school in Riverton City equals 2 plus the number of elementary school in Riverton City. The number of elementary school in Riverton City exists, and its number is greater than 0. The number of elementary school in Riverton City exists, and its number is greater than 0. The number of elementary school in Shoreline City equals the sum of the number of regional medical school in Riverton City, the number of elementary school in Riverton City, the total number of schools in Riverton City, and the number of private middle school in Riverton City. The total number of schools in Shoreline City equals 48.

Question: How many elementary school does Riverton City have?

Solution: The question is difficult, so we use equations to solve it. Define elementary school in Riverton City as x; we don't know its value yet but will find it out later. Define regional medical school in Riverton City as E; Z = x = x; so E = 2 + Z = 2 + x = x + 2. Define private middle school in Riverton City as G; V = x = x; so G = 1 + V = 1 + x = x + 1. Define total number of schools in Riverton City as J; L = x + G = x + x + 1 = 2*x + 1; so J = L + E = 2*x + 1 + x + 2 = 3*x + 3. Define regional medical school in Shoreline City as p; I = J + x = 3*x + 3 + x = 4*x + 3; so p = I + G = 4*x + 3 + x + 1 = 5*x + 4. Define elementary school in Shoreline City as W; w = E + x = x + 2 + x = 2*x + 2; s = w + J = 2*x + 2 + 3*x + 3 = 5*x + 5; so W = s + G = 5*x + 5 + x + 1 = 6*x + 6. Define private middle school in Shoreline City as f; so f = E = x + 2. Define total number of schools in Shoreline City as O; k = W + f = 6*x + 6 + x + 2 = 7*x + 8; so O = k + p = 7*x + 8 + 5*x + 4 = 12*x + 12. We know O = 48, so we have 12*x + 12 = 48 and we solve it through the following steps: Simplifying the equation: 12*x + 12 = 48. Move all terms to one side: 12*x - 36 = 0. Isolate the term with x: 12*x = 36. Divide both sides by 12: x = 36 / 12. Solution: x = 3. Answer: 3.


The number of adult blue jay in Maple Creek exists, and its number is greater than 0. The number of adult blue jay in Maple Creek exists, and its number is greater than 0. The number of adult owl in Maple Creek equals 3 times the number of adult blue jay in Maple Creek. The number of adult eagle in Cedar Valley equals the sum of the number of adult owl in Cedar Valley, the total number of adult animals in Maple Creek, and the number of adult blue jay in Cedar Valley. The total number of adult animals in Cedar Valley equals 172. The number of adult owl in Cedar Valley equals the sum of the number of adult blue jay in Cedar Valley, the number of adult owl in Maple Creek, the total number of adult animals in Maple Creek, and the number of adult blue jay in Maple Creek.

Question: How many adult blue jay does Maple Creek have?

Solution: The question is difficult, so we use equations to solve it. Define adult blue jay in Maple Creek as x; we don't know its value yet but will find it out later. Define adult owl in Maple Creek as a; n = x = x; so a = 3 * n = (3) * (x) = 3*x. Define total number of adult animals in Maple Creek as A; so A = a + x = 3*x + x = 4*x. Define adult blue jay in Cedar Valley as R; G = A + x = 4*x + x = 5*x; h = G + a = 5*x + 3*x = 8*x; so R = 4 + h = 4 + 8*x = 8*x + 4. Define adult owl in Cedar Valley as q; j = R + a = 8*x + 4 + 3*x = 11*x + 4; E = j + A = 11*x + 4 + 4*x = 15*x + 4; so q = E + x = 15*x + 4 + x = 16*x + 4. Define adult eagle in Cedar Valley as B; t = q + A = 16*x + 4 + 4*x = 20*x + 4; so B = t + R = 20*x + 4 + 8*x + 4 = 28*x + 8. Define total number of adult animals in Cedar Valley as K; Z = q + B = 16*x + 4 + 28*x + 8 = 44*x + 12; so K = Z + R = 44*x + 12 + 8*x + 4 = 52*x + 16. We know K = 172, so we have 52*x + 16 = 172 and we solve it through the following steps: Simplifying the equation: 52*x + 16 = 172. Move all terms to one side: 52*x - 156 = 0. Isolate the term with x: 52*x = 156. Divide both sides by 52: x = 156 / 52. Solution: x = 3. Answer: 3.


The total number of adult animals in Maple Creek equals 31. The number of adult racoon in Oakridge Riverside exists, and its number is greater than 0. The number of adult fox in Beverly Forest equals 4 plus the total number of adult animals in Oakridge Riverside. The number of adult bear in Pine Ridge equals 4 plus the sum of the total number of adult animals in Beverly Forest, the total number of adult animals in Oakridge Riverside, and the number of adult racoon in Oakridge Riverside. The number of adult deer in Cedar Valley equals the sum of the total number of adult animals in Pine Ridge, the total number of adult animals in Oakridge Riverside, the number of adult bear in Pine Ridge, and the total number of adult animals in Beverly Forest. The number of adult wolf in Maple Creek equals 1 plus the difference between the total number of adult animals in Cedar Valley and the number of adult fox in Beverly Forest.

Question: How many adult racoon does Oakridge Riverside have?

Solution: The question is difficult, so we use equations to solve it. Define adult racoon in Oakridge Riverside as x; we don't know its value yet but will find it out later. Define total number of adult animals in Oakridge Riverside as N; so N = x = x. Define adult fox in Beverly Forest as L; A = N = x; so L = 4 + A = 4 + x = x + 4. Define total number of adult animals in Beverly Forest as B; so B = L = x + 4. Define adult bear in Pine Ridge as M; t = B + N = x + 4 + x = 2*x + 4; k = t + x = 2*x + 4 + x = 3*x + 4; so M = 4 + k = 4 + 3*x + 4 = 3*x + 8. Define total number of adult animals in Pine Ridge as j; so j = M = 3*x + 8. Define adult deer in Cedar Valley as Q; f = j + N = 3*x + 8 + x = 4*x + 8; H = f + M = 4*x + 8 + 3*x + 8 = 7*x + 16; so Q = H + B = 7*x + 16 + x + 4 = 8*x + 20. Define total number of adult animals in Cedar Valley as l; so l = Q = 8*x + 20. Define adult wolf in Maple Creek as E; Z = l - L = (8*x + 20) - (x + 4) = 7*x + 16; so E = 1 + Z = 1 + 7*x + 16 = 7*x + 17. Define total number of adult animals in Maple Creek as V; so V = E = 7*x + 17. We know V = 31, so we have 7*x + 17 = 31 and we solve it through the following steps: Simplifying the equation: 7*x + 17 = 31. Move all terms to one side: 7*x - 14 = 0. Isolate the term with x: 7*x = 14. Divide both sides by 7: x = 14 / 7. Solution: x = 2. Answer: 2.


The total number of movies in Festival de Clairmont equals 45. The number of solemn period drama in Cinéma de Montreval exists, and its number is greater than 0. The number of solemn period drama in Rêves de Belleville equals the number of solemn period drama in Cinéma de Montreval. The number of intense detective thriller in Cinéma de Montreval equals the number of intense detective thriller in Festival de Clairmont. The number of intense detective thriller in Festival de Saint-Rivage equals 3 plus the difference between the number of solemn period drama in Rêves de Belleville and the number of solemn period drama in Cinéma de Montreval. The number of intense detective thriller in Festival de Clairmont equals 4 plus the sum of the number of calm road movie in Festival de Saint-Rivage, the number of intense detective thriller in Festival de Saint-Rivage, and the number of solemn period drama in Cinéma de Montreval. The number of solemn period drama in Cinéma de Montreval exists, and its number is greater than 0. The number of calm road movie in Festival de Clairmont equals the sum of the number of intense detective thriller in Festival de Clairmont, the number of solemn period drama in Cinéma de Montreval, the number of intense detective thriller in Festival de Saint-Rivage, and the number of solemn period drama in Rêves de Belleville. The number of upbeat metropolis comedy in Cinéma de Montreval equals 3.

Question: How many solemn period drama does Cinéma de Montreval have?

Solution: The question is difficult, so we use equations to solve it. Define solemn period drama in Cinéma de Montreval as x; we don't know its value yet but will find it out later. Define solemn period drama in Rêves de Belleville as l; so l = x = x. Define intense detective thriller in Festival de Saint-Rivage as B; e = l - x = (x) - (x) = 0; so B = 3 + e = 3 + 0 = 3. Define calm road movie in Festival de Saint-Rivage as P; K = B + x = 3 + x = x + 3; J = K + l = x + 3 + x = 2*x + 3; so P = 3 + J = 3 + 2*x + 3 = 2*x + 6. Define intense detective thriller in Festival de Clairmont as c; f = P + B = 2*x + 6 + 3 = 2*x + 9; F = f + x = 2*x + 9 + x = 3*x + 9; so c = 4 + F = 4 + 3*x + 9 = 3*x + 13. Define calm road movie in Festival de Clairmont as I; C = c + x = 3*x + 13 + x = 4*x + 13; j = C + B = 4*x + 13 + 3 = 4*x + 16; so I = j + l = 4*x + 16 + x = 5*x + 16. Define total number of movies in Festival de Clairmont as R; so R = c + I = 3*x + 13 + 5*x + 16 = 8*x + 29. We know R = 45, so we have 8*x + 29 = 45 and we solve it through the following steps: Simplifying the equation: 8*x + 29 = 45. Move all terms to one side: 8*x - 16 = 0. Isolate the term with x: 8*x = 16. Divide both sides by 8: x = 16 / 8. Solution: x = 2. Answer: 2.


The number of adult eagle in Maple Creek equals 4 plus the sum of the number of adult parrot in Maple Creek and the number of adult owl in Oakridge Riverside. The total number of adult animals in Beverly Forest equals 34. The number of adult parrot in Beverly Forest equals 3 plus the sum of the number of adult eagle in Maple Creek and the total number of adult animals in Oakridge Riverside. The number of adult owl in Cedar Valley equals 3 plus the number of adult eagle in Beverly Forest. The number of adult owl in Oakridge Riverside exists, and its number is greater than 0. The number of adult eagle in Cedar Valley equals 3. The number of adult owl in Oakridge Riverside exists, and its number is greater than 0. The number of adult parrot in Maple Creek equals 2 plus the difference between the total number of adult animals in Oakridge Riverside and the number of adult owl in Oakridge Riverside.

Question: How many adult owl does Oakridge Riverside have?

Solution: The question is difficult, so we use equations to solve it. Define adult owl in Oakridge Riverside as x; we don't know its value yet but will find it out later. Define total number of adult animals in Oakridge Riverside as K; so K = x = x. Define adult parrot in Maple Creek as a; t = K - x = (x) - (x) = 0; so a = 2 + t = 2 + 0 = 2. Define adult eagle in Maple Creek as G; p = a + x = 2 + x = x + 2; so G = 4 + p = 4 + x + 2 = x + 6. Define total number of adult animals in Maple Creek as d; so d = a + G = 2 + x + 6 = x + 8. Define adult eagle in Beverly Forest as P; f = d + a = x + 8 + 2 = x + 10; J = f + x = x + 10 + x = 2*x + 10; so P = J + K = 2*x + 10 + x = 3*x + 10. Define adult parrot in Beverly Forest as r; n = G + K = x + 6 + x = 2*x + 6; so r = 3 + n = 3 + 2*x + 6 = 2*x + 9. Define total number of adult animals in Beverly Forest as M; so M = r + P = 2*x + 9 + 3*x + 10 = 5*x + 19. We know M = 34, so we have 5*x + 19 = 34 and we solve it through the following steps: Simplifying the equation: 5*x + 19 = 34. Move all terms to one side: 5*x - 15 = 0. Isolate the term with x: 5*x = 15. Divide both sides by 5: x = 15 / 5. Solution: x = 3. Answer: 3.


The total number of schools in Shoreline City equals 54. The number of private middle school in Shoreline City equals 4 plus the difference between the total number of schools in Clearwater Bay and the number of regional medical school in Riverton City. The number of private middle school in Clearwater Bay equals the number of regional medical school in Clearwater Bay. The number of regional medical school in Clearwater Bay equals 3. The number of regional medical school in Shoreline City equals 2 times the sum of the total number of schools in Riverton City, the total number of schools in Clearwater Bay, and the number of regional medical school in Clearwater Bay. The number of private middle school in Riverton City equals the sum of the total number of schools in Clearwater Bay, the number of private middle school in Clearwater Bay, and the number of regional medical school in Clearwater Bay. The number of regional medical school in Riverton City exists, and its number is greater than 0.

Question: How many regional medical school does Riverton City have?

Solution: The question is difficult, so we use equations to solve it. Define regional medical school in Clearwater Bay as c; so c = 3. Define private middle school in Clearwater Bay as W; so W = c = 3. Define total number of schools in Clearwater Bay as S; so S = c + W = 3 + 3 = 6. Define private middle school in Riverton City as o; m = S + W = 6 + 3 = 9; so o = m + c = 9 + 3 = 12. Define regional medical school in Riverton City as x; we don't know its value yet but will find it out later. Define total number of schools in Riverton City as P; so P = x + o = x + 12 = x + 12. Define regional medical school in Shoreline City as M; k = P + S = x + 12 + 6 = x + 18; j = k + c = x + 18 + 3 = x + 21; so M = 2 * j = (2) * (x + 21) = 2*x + 42. Define private middle school in Shoreline City as z; t = S - x = (6) - (x) = 6 - x; so z = 4 + t = 4 + 6 - x = 10 - x. Define total number of schools in Shoreline City as w; so w = M + z = 2*x + 42 + 10 - x = x + 52. We know w = 54, so we have x + 52 = 54 and we solve it through the following steps: Simplifying the equation: x + 52 = 54. Move all terms to one side: x - 2 = 0. Isolate the term with x: 1*x = 2. Divide both sides by 1: x = 2 / 1. Solution: x = 2. Answer: 2.


The number of upbeat metropolis comedy in Festival Lumière de Valmont equals the sum of the number of calm road movie in Rêves de Belleville, the number of solemn period drama in Festival Lumière de Valmont, and the number of calm road movie in Cinéma de Montreval. The number of calm road movie in Cinéma de Montreval exists, and its number is greater than 0. The number of calm road movie in Rêves de Belleville equals 4 plus the difference between the number of solemn period drama in Festival Lumière de Valmont and the total number of movies in Cinéma de Montreval. The total number of movies in Rêves de Belleville equals 63. The number of solemn period drama in Festival Lumière de Valmont equals 3 plus the sum of the total number of movies in Cinéma de Montreval and the number of calm road movie in Cinéma de Montreval. The number of solemn period drama in Rêves de Belleville equals the total number of movies in Festival Lumière de Valmont. The number of upbeat metropolis comedy in Rêves de Belleville equals the difference between the total number of movies in Festival Lumière de Valmont and the number of solemn period drama in Festival Lumière de Valmont.

Question: How many calm road movie does Cinéma de Montreval have?

Solution: The question is difficult, so we use equations to solve it. Define calm road movie in Cinéma de Montreval as x; we don't know its value yet but will find it out later. Define total number of movies in Cinéma de Montreval as Q; so Q = x = x. Define solemn period drama in Festival Lumière de Valmont as Y; D = Q + x = x + x = 2*x; so Y = 3 + D = 3 + 2*x = 2*x + 3. Define calm road movie in Rêves de Belleville as M; H = Y - Q = (2*x + 3) - (x) = x + 3; so M = 4 + H = 4 + x + 3 = x + 7. Define upbeat metropolis comedy in Festival Lumière de Valmont as g; R = M + Y = x + 7 + 2*x + 3 = 3*x + 10; so g = R + x = 3*x + 10 + x = 4*x + 10. Define total number of movies in Festival Lumière de Valmont as C; so C = g + Y = 4*x + 10 + 2*x + 3 = 6*x + 13. Define solemn period drama in Rêves de Belleville as I; so I = C = 6*x + 13. Define upbeat metropolis comedy in Rêves de Belleville as h; so h = C - Y = (6*x + 13) - (2*x + 3) = 4*x + 10. Define total number of movies in Rêves de Belleville as L; i = M + h = x + 7 + 4*x + 10 = 5*x + 17; so L = i + I = 5*x + 17 + 6*x + 13 = 11*x + 30. We know L = 63, so we have 11*x + 30 = 63 and we solve it through the following steps: Simplifying the equation: 11*x + 30 = 63. Move all terms to one side: 11*x - 33 = 0. Isolate the term with x: 11*x = 33. Divide both sides by 11: x = 33 / 11. Solution: x = 3. Answer: 3.


The number of adult racoon in Jefferson Circus exists, and its number is greater than 0. The number of adult racoon in Jefferson Circus exists, and its number is greater than 0. The number of adult fox in Jefferson Circus equals 3 plus the number of adult racoon in Jefferson Circus. The total number of adult animals in Mayer Aquarium equals 53. The number of adult racoon in Jefferson Circus exists, and its number is greater than 0. The number of adult bear in Bundle Ranch equals 2 times the difference between the total number of adult animals in Jefferson Circus and the number of adult racoon in Jefferson Circus.

Question: How many adult racoon does Jefferson Circus have?

Solution: The question is difficult, so we use equations to solve it. Define adult racoon in Jefferson Circus as x; we don't know its value yet but will find it out later. Define adult fox in Jefferson Circus as n; e = x = x; so n = 3 + e = 3 + x = x + 3. Define total number of adult animals in Jefferson Circus as D; so D = n + x = x + 3 + x = 2*x + 3. Define adult bear in Bundle Ranch as K; O = D - x = (2*x + 3) - (x) = x + 3; so K = 2 * O = (2) * (x + 3) = 2*x + 6. Define total number of adult animals in Bundle Ranch as G; so G = K = 2*x + 6. Define adult deer in Hamilton Farm as g; y = G + x = 2*x + 6 + x = 3*x + 6; l = y + K = 3*x + 6 + 2*x + 6 = 5*x + 12; so g = 2 + l = 2 + 5*x + 12 = 5*x + 14. Define total number of adult animals in Hamilton Farm as S; so S = g = 5*x + 14. Define adult wolf in Mayer Aquarium as k; z = S + D = 5*x + 14 + 2*x + 3 = 7*x + 17; j = z + x = 7*x + 17 + x = 8*x + 17; so k = j + K = 8*x + 17 + 2*x + 6 = 10*x + 23. Define total number of adult animals in Mayer Aquarium as P; so P = k = 10*x + 23. We know P = 53, so we have 10*x + 23 = 53 and we solve it through the following steps: Simplifying the equation: 10*x + 23 = 53. Move all terms to one side: 10*x - 30 = 0. Isolate the term with x: 10*x = 30. Divide both sides by 10: x = 30 / 10. Solution: x = 3. Answer: 3.


The number of regional medical school in Shoreline City exists, and its number is greater than 0. The number of regional medical school in Riverton City equals 3 plus the number of culinarian school in Riverton City. The total number of schools in Oakbridge City equals 98. The number of regional medical school in Oakbridge City equals the sum of the total number of schools in Shoreline City, the number of regional medical school in Riverton City, and the number of regional medical school in Shoreline City. The number of culinarian school in Riverton City equals 3. The number of elementary school in Shoreline City equals 4 times the sum of the number of culinarian school in Riverton City, the number of regional medical school in Riverton City, and the total number of schools in Riverton City. The number of culinarian school in Shoreline City equals 2 times the total number of schools in Riverton City.

Question: How many regional medical school does Shoreline City have?

Solution: The question is difficult, so we use equations to solve it. Define culinarian school in Riverton City as A; so A = 3. Define regional medical school in Riverton City as D; Y = A = 3; so D = 3 + Y = 3 + 3 = 6. Define total number of schools in Riverton City as r; so r = A + D = 3 + 6 = 9. Define elementary school in Shoreline City as w; J = A + D = 3 + 6 = 9; V = J + r = 9 + 9 = 18; so w = 4 * V = (4) * (18) = 72. Define culinarian school in Shoreline City as Z; c = r = 9; so Z = 2 * c = (2) * (9) = 18. Define regional medical school in Shoreline City as x; we don't know its value yet but will find it out later. Define total number of schools in Shoreline City as H; d = Z + x = 18 + x = x + 18; so H = d + w = x + 18 + 72 = x + 90. Define regional medical school in Oakbridge City as Q; T = H + D = x + 90 + 6 = x + 96; so Q = T + x = x + 96 + x = 2*x + 96. Define total number of schools in Oakbridge City as k; so k = Q = 2*x + 96. We know k = 98, so we have 2*x + 96 = 98 and we solve it through the following steps: Simplifying the equation: 2*x + 96 = 98. Move all terms to one side: 2*x - 2 = 0. Isolate the term with x: 2*x = 2. Divide both sides by 2: x = 2 / 2. Solution: x = 1. Answer: 1.


The number of adult parrot in Cedar Valley exists, and its number is greater than 0. The number of adult crow in Cedar Valley equals 4 times the number of adult parrot in Cedar Valley. The number of adult crow in Beverly Forest equals the sum of the total number of adult animals in Oakridge Riverside, the number of adult blue jay in Oakridge Riverside, and the number of adult parrot in Cedar Valley. The total number of adult animals in Pine Ridge equals 97. The number of adult parrot in Cedar Valley exists, and its number is greater than 0. The number of adult parrot in Cedar Valley exists, and its number is greater than 0.

Question: How many adult parrot does Cedar Valley have?

Solution: The question is difficult, so we use equations to solve it. Define adult parrot in Cedar Valley as x; we don't know its value yet but will find it out later. Define adult crow in Cedar Valley as R; z = x = x; so R = 4 * z = (4) * (x) = 4*x. Define total number of adult animals in Cedar Valley as y; so y = R + x = 4*x + x = 5*x. Define adult blue jay in Oakridge Riverside as a; A = y + x = 5*x + x = 6*x; t = A + R = 6*x + 4*x = 10*x; so a = 2 + t = 2 + 10*x = 10*x + 2. Define total number of adult animals in Oakridge Riverside as P; so P = a = 10*x + 2. Define adult crow in Beverly Forest as F; G = P + a = 10*x + 2 + 10*x + 2 = 20*x + 4; so F = G + x = 20*x + 4 + x = 21*x + 4. Define total number of adult animals in Beverly Forest as I; so I = F = 21*x + 4. Define adult eagle in Pine Ridge as u; c = I + R = 21*x + 4 + 4*x = 25*x + 4; S = c + x = 25*x + 4 + x = 26*x + 4; so u = S + y = 26*x + 4 + 5*x = 31*x + 4. Define total number of adult animals in Pine Ridge as e; so e = u = 31*x + 4. We know e = 97, so we have 31*x + 4 = 97 and we solve it through the following steps: Simplifying the equation: 31*x + 4 = 97. Move all terms to one side: 31*x - 93 = 0. Isolate the term with x: 31*x = 93. Divide both sides by 31: x = 93 / 31. Solution: x = 3. Answer: 3.


The number of public highschool in Glenfield City equals the number of elementary school in Glenfield City. The number of public highschool in Brightford equals the difference between the number of private middle school in Evervale City and the number of public highschool in Evervale City. The number of elementary school in Glenfield City exists, and its number is greater than 0. The number of private middle school in Evervale City equals 4 plus the number of private middle school in Brightford. The number of public highschool in Westhaven City equals the sum of the number of private middle school in Glenfield City, the total number of schools in Glenfield City, the number of public highschool in Glenfield City, and the number of elementary school in Glenfield City. The number of private middle school in Glenfield City equals the number of public highschool in Glenfield City. The number of elementary school in Glenfield City exists, and its number is greater than 0. The number of private middle school in Brightford equals the number of public highschool in Evervale City. The number of public highschool in Evervale City equals 4. The number of elementary school in Westhaven City equals 3 plus the number of private middle school in Glenfield City. The total number of schools in Westhaven City equals 54. The number of elementary school in Evervale City equals 4.

Question: How many elementary school does Glenfield City have?

Solution: The question is difficult, so we use equations to solve it. Define elementary school in Glenfield City as x; we don't know its value yet but will find it out later. Define public highschool in Glenfield City as P; so P = x = x. Define private middle school in Glenfield City as T; so T = P = x. Define total number of schools in Glenfield City as G; Z = x + T = x + x = 2*x; so G = Z + P = 2*x + x = 3*x. Define private middle school in Westhaven City as B; k = G + x = 3*x + x = 4*x; U = k + P = 4*x + x = 5*x; so B = 3 + U = 3 + 5*x = 5*x + 3. Define elementary school in Westhaven City as t; e = T = x; so t = 3 + e = 3 + x = x + 3. Define public highschool in Westhaven City as Q; d = T + G = x + 3*x = 4*x; E = d + P = 4*x + x = 5*x; so Q = E + x = 5*x + x = 6*x. Define total number of schools in Westhaven City as O; v = t + B = x + 3 + 5*x + 3 = 6*x + 6; so O = v + Q = 6*x + 6 + 6*x = 12*x + 6. We know O = 54, so we have 12*x + 6 = 54 and we solve it through the following steps: Simplifying the equation: 12*x + 6 = 54. Move all terms to one side: 12*x - 48 = 0. Isolate the term with x: 12*x = 48. Divide both sides by 12: x = 48 / 12. Solution: x = 4. Answer: 4.


The number of solemn period drama in Festival de Clairmont equals 2. The number of calm road movie in Rêves de Belleville equals 4. The number of futuristic sci-fi movie in Festival de Clairmont equals the number of calm road movie in Festival de Clairmont. The number of solemn period drama in Festival de Saint-Rivage equals the number of futuristic sci-fi movie in Festival de Saint-Rivage. The number of solemn period drama in Cinéma de Montreval equals 3 plus the sum of the number of solemn period drama in Festival de Saint-Rivage and the number of futuristic sci-fi movie in Festival de Saint-Rivage. The number of futuristic sci-fi movie in Festival de Saint-Rivage equals 3. The number of calm road movie in Festival de Saint-Rivage equals 3 plus the sum of the total number of movies in Cinéma de Montreval, the number of futuristic sci-fi movie in Cinéma de Montreval, and the number of solemn period drama in Cinéma de Montreval. The number of calm road movie in Festival de Clairmont equals 4 plus the number of calm road movie in Rêves de Belleville. The number of futuristic sci-fi movie in Cinéma de Montreval equals the sum of the number of solemn period drama in Cinéma de Montreval, the number of solemn period drama in Festival de Saint-Rivage, and the number of futuristic sci-fi movie in Festival de Saint-Rivage. The number of calm road movie in Cinéma de Montreval equals 3 plus the sum of the number of solemn period drama in Cinéma de Montreval, the number of futuristic sci-fi movie in Festival de Saint-Rivage, and the number of solemn period drama in Festival de Saint-Rivage. The number of solemn period drama in Rêves de Belleville equals 2 plus the sum of the number of futuristic sci-fi movie in Cinéma de Montreval and the number of futuristic sci-fi movie in Festival de Saint-Rivage.

Question: How many calm road movie does Festival de Saint-Rivage have?

Solution: Define futuristic sci-fi movie in Festival de Saint-Rivage as G; so G = 3. Define solemn period drama in Festival de Saint-Rivage as D; so D = G = 3. Define solemn period drama in Cinéma de Montreval as i; n = D + G = 3 + 3 = 6; so i = 3 + n = 3 + 6 = 9. Define calm road movie in Cinéma de Montreval as k; X = i + G = 9 + 3 = 12; H = X + D = 12 + 3 = 15; so k = 3 + H = 3 + 15 = 18. Define futuristic sci-fi movie in Cinéma de Montreval as f; N = i + D = 9 + 3 = 12; so f = N + G = 12 + 3 = 15. Define total number of movies in Cinéma de Montreval as A; l = i + k = 9 + 18 = 27; so A = l + f = 27 + 15 = 42. Define calm road movie in Festival de Saint-Rivage as Q; t = A + f = 42 + 15 = 57; d = t + i = 57 + 9 = 66; so Q = 3 + d = 3 + 66 = 69. Answer: 69.


The number of elementary school in Evervale City equals 3 times the number of public highschool in Evervale City. The number of culinarian school in Brightford equals the total number of schools in Westhaven City. The number of elementary school in Brightford equals 1 plus the number of culinarian school in Evervale City. The number of culinarian school in Westhaven City exists, and its number is greater than 0. The number of culinarian school in Evervale City equals 1 plus the number of public highschool in Evervale City. The number of public highschool in Evervale City equals 4. The number of public highschool in Westhaven City equals the number of public highschool in Evervale City. The number of elementary school in Westhaven City equals the total number of schools in Evervale City. The total number of schools in Brightford equals 34.

Question: How many culinarian school does Westhaven City have?

Solution: The question is difficult, so we use equations to solve it. Define public highschool in Evervale City as g; so g = 4. Define elementary school in Evervale City as c; p = g = 4; so c = 3 * p = (3) * (4) = 12. Define culinarian school in Evervale City as y; K = g = 4; so y = 1 + K = 1 + 4 = 5. Define total number of schools in Evervale City as h; B = c + y = 12 + 5 = 17; so h = B + g = 17 + 4 = 21. Define elementary school in Westhaven City as a; so a = h = 21. Define public highschool in Westhaven City as S; so S = g = 4. Define culinarian school in Westhaven City as x; we don't know its value yet but will find it out later. Define total number of schools in Westhaven City as P; U = a + x = 21 + x = x + 21; so P = U + S = x + 21 + 4 = x + 25. Define culinarian school in Brightford as F; so F = P = x + 25. Define elementary school in Brightford as b; I = y = 5; so b = 1 + I = 1 + 5 = 6. Define total number of schools in Brightford as L; so L = b + F = 6 + x + 25 = x + 31. We know L = 34, so we have x + 31 = 34 and we solve it through the following steps: Simplifying the equation: x + 31 = 34. Move all terms to one side: x - 3 = 0. Isolate the term with x: 1*x = 3. Divide both sides by 1: x = 3 / 1. Solution: x = 3. Answer: 3.


The number of intense detective thriller in Festival Lumière de Valmont exists, and its number is greater than 0. The number of intense detective thriller in Festival de Saint-Rivage equals 4 plus the total number of movies in Festival Lumière de Valmont. The number of futuristic sci-fi movie in Festival Lumière de Valmont equals 2 plus the number of intense detective thriller in Festival Lumière de Valmont. The number of calm road movie in Rêves de Belleville equals the sum of the number of intense detective thriller in Festival de Saint-Rivage, the total number of movies in Festival Lumière de Valmont, and the total number of movies in Festival de Saint-Rivage. The total number of movies in Cinéma de Montreval equals 116. The number of solemn period drama in Cinéma de Montreval equals the sum of the total number of movies in Rêves de Belleville and the number of calm road movie in Rêves de Belleville. The number of calm road movie in Cinéma de Montreval equals 2 times the sum of the total number of movies in Festival de Saint-Rivage and the number of futuristic sci-fi movie in Festival Lumière de Valmont.

Question: How many intense detective thriller does Festival Lumière de Valmont have?

Solution: The question is difficult, so we use equations to solve it. Define intense detective thriller in Festival Lumière de Valmont as x; we don't know its value yet but will find it out later. Define futuristic sci-fi movie in Festival Lumière de Valmont as M; e = x = x; so M = 2 + e = 2 + x = x + 2. Define total number of movies in Festival Lumière de Valmont as w; so w = M + x = x + 2 + x = 2*x + 2. Define intense detective thriller in Festival de Saint-Rivage as I; Q = w = 2*x + 2; so I = 4 + Q = 4 + 2*x + 2 = 2*x + 6. Define total number of movies in Festival de Saint-Rivage as F; so F = I = 2*x + 6. Define calm road movie in Cinéma de Montreval as d; S = F + M = 2*x + 6 + x + 2 = 3*x + 8; so d = 2 * S = (2) * (3*x + 8) = 6*x + 16. Define calm road movie in Rêves de Belleville as c; n = I + w = 2*x + 6 + 2*x + 2 = 4*x + 8; so c = n + F = 4*x + 8 + 2*x + 6 = 6*x + 14. Define total number of movies in Rêves de Belleville as p; so p = c = 6*x + 14. Define solemn period drama in Cinéma de Montreval as Y; so Y = p + c = 6*x + 14 + 6*x + 14 = 12*x + 28. Define total number of movies in Cinéma de Montreval as k; so k = Y + d = 12*x + 28 + 6*x + 16 = 18*x + 44. We know k = 116, so we have 18*x + 44 = 116 and we solve it through the following steps: Simplifying the equation: 18*x + 44 = 116. Move all terms to one side: 18*x - 72 = 0. Isolate the term with x: 18*x = 72. Divide both sides by 18: x = 72 / 18. Solution: x = 4. Answer: 4.


The number of solemn period drama in Rêves de Belleville equals 1 plus the sum of the number of futuristic sci-fi movie in Festival Lumière de Valmont, the number of upbeat metropolis comedy in Cinéma de Montreval, and the number of futuristic sci-fi movie in Cinéma de Montreval. The number of futuristic sci-fi movie in Festival de Clairmont equals 3 plus the number of upbeat metropolis comedy in Festival de Clairmont. The number of intense detective thriller in Festival Lumière de Valmont equals the number of upbeat metropolis comedy in Festival de Saint-Rivage. The number of upbeat metropolis comedy in Rêves de Belleville equals 2 times the difference between the number of calm road movie in Festival de Saint-Rivage and the number of solemn period drama in Rêves de Belleville. The number of solemn period drama in Cinéma de Montreval equals 2 plus the sum of the number of calm road movie in Festival Lumière de Valmont, the number of futuristic sci-fi movie in Festival de Saint-Rivage, and the number of intense detective thriller in Festival de Saint-Rivage. The number of intense detective thriller in Festival de Saint-Rivage equals 4. The number of calm road movie in Festival Lumière de Valmont equals the sum of the number of futuristic sci-fi movie in Festival Lumière de Valmont, the number of upbeat metropolis comedy in Festival de Saint-Rivage, and the number of upbeat metropolis comedy in Festival Lumière de Valmont. The number of upbeat metropolis comedy in Festival Lumière de Valmont equals 4 times the number of upbeat metropolis comedy in Festival de Saint-Rivage. The number of solemn period drama in Festival de Clairmont equals 4 plus the sum of the number of futuristic sci-fi movie in Cinéma de Montreval and the number of upbeat metropolis comedy in Festival de Clairmont. The number of upbeat metropolis comedy in Cinéma de Montreval equals the sum of the number of calm road movie in Festival de Clairmont, the number of futuristic sci-fi movie in Festival de Clairmont, the number of futuristic sci-fi movie in Cinéma de Montreval, and the number of upbeat metropolis comedy in Festival de Clairmont. The number of futuristic sci-fi movie in Festival Lumière de Valmont equals 3 plus the number of upbeat metropolis comedy in Festival de Saint-Rivage. The number of calm road movie in Festival de Saint-Rivage equals the sum of the total number of movies in Festival Lumière de Valmont, the number of upbeat metropolis comedy in Festival de Saint-Rivage, and the number of upbeat metropolis comedy in Festival Lumière de Valmont. The number of upbeat metropolis comedy in Festival de Clairmont equals 2. The number of futuristic sci-fi movie in Festival de Saint-Rivage equals the sum of the number of calm road movie in Festival de Saint-Rivage, the number of calm road movie in Festival Lumière de Valmont, and the number of upbeat metropolis comedy in Festival Lumière de Valmont. The number of upbeat metropolis comedy in Festival de Saint-Rivage equals 4. The number of futuristic sci-fi movie in Cinéma de Montreval equals the number of futuristic sci-fi movie in Festival de Clairmont. The number of calm road movie in Festival de Clairmont equals 1 plus the number of intense detective thriller in Festival de Saint-Rivage.

Question: How many futuristic sci-fi movie does Festival de Saint-Rivage have?

Solution: Define upbeat metropolis comedy in Festival de Saint-Rivage as L; so L = 4. Define futuristic sci-fi movie in Festival Lumière de Valmont as G; a = L = 4; so G = 3 + a = 3 + 4 = 7. Define upbeat metropolis comedy in Festival Lumière de Valmont as s; V = L = 4; so s = 4 * V = 4 * 4 = 16. Define calm road movie in Festival Lumière de Valmont as j; J = G + L = 7 + 4 = 11; so j = J + s = 11 + 16 = 27. Define intense detective thriller in Festival Lumière de Valmont as S; so S = L = 4. Define total number of movies in Festival Lumière de Valmont as N; B = G + j = 7 + 27 = 34; x = B + s = 34 + 16 = 50; so N = x + S = 50 + 4 = 54. Define calm road movie in Festival de Saint-Rivage as f; o = N + L = 54 + 4 = 58; so f = o + s = 58 + 16 = 74. Define futuristic sci-fi movie in Festival de Saint-Rivage as c; U = f + j = 74 + 27 = 101; so c = U + s = 101 + 16 = 117. Answer: 117.


The number of adult deer in Maple Creek equals 4 times the number of adult wolf in Beverly Forest. The number of adult racoon in Maple Creek equals 2. The number of adult deer in Beverly Forest equals 2 plus the sum of the total number of adult animals in Oakridge Riverside, the number of adult deer in Oakridge Riverside, and the total number of adult animals in Maple Creek. The number of adult wolf in Maple Creek equals 1 plus the number of adult wolf in Beverly Forest. The number of adult deer in Oakridge Riverside equals 2 times the difference between the total number of adult animals in Maple Creek and the number of adult deer in Maple Creek. The number of adult wolf in Beverly Forest equals 1.

Question: What is the total number of adult animals in Beverly Forest?

Solution: Define adult wolf in Beverly Forest as L; so L = 1. Define adult wolf in Maple Creek as h; d = L = 1; so h = 1 + d = 1 + 1 = 2. Define adult deer in Maple Creek as D; W = L = 1; so D = 4 * W = 4 * 1 = 4. Define adult racoon in Maple Creek as P; so P = 2. Define total number of adult animals in Maple Creek as J; R = P + D = 2 + 4 = 6; so J = R + h = 6 + 2 = 8. Define adult deer in Oakridge Riverside as A; I = J - D = 8 - 4 = 4; so A = 2 * I = 2 * 4 = 8. Define total number of adult animals in Oakridge Riverside as Z; so Z = A = 8. Define adult deer in Beverly Forest as u; t = Z + A = 8 + 8 = 16; g = t + J = 16 + 8 = 24; so u = 2 + g = 2 + 24 = 26. Define total number of adult animals in Beverly Forest as i; so i = u + L = 26 + 1 = 27. Answer: 27.


The number of adult wolf in Bundle Ranch equals 1. The number of adult wolf in Hamilton Farm equals the sum of the total number of adult animals in Bundle Ranch, the number of adult wolf in Bundle Ranch, and the number of adult deer in Bundle Ranch. The number of adult racoon in Mayer Aquarium equals the number of adult bear in Jefferson Circus. The number of adult wolf in Mayer Aquarium equals the total number of adult animals in Jefferson Circus. The number of adult bear in Jefferson Circus equals 1 plus the sum of the total number of adult animals in Hamilton Farm, the number of adult wolf in Bundle Ranch, and the total number of adult animals in Bundle Ranch. The number of adult racoon in Hamilton Farm equals 4. The number of adult deer in Bundle Ranch equals the number of adult wolf in Bundle Ranch.

Question: What is the total number of adult animals in Mayer Aquarium?

Solution: Define adult wolf in Bundle Ranch as y; so y = 1. Define adult deer in Bundle Ranch as T; so T = y = 1. Define total number of adult animals in Bundle Ranch as a; so a = T + y = 1 + 1 = 2. Define adult wolf in Hamilton Farm as b; u = a + y = 2 + 1 = 3; so b = u + T = 3 + 1 = 4. Define adult racoon in Hamilton Farm as m; so m = 4. Define total number of adult animals in Hamilton Farm as A; so A = b + m = 4 + 4 = 8. Define adult bear in Jefferson Circus as g; C = A + y = 8 + 1 = 9; c = C + a = 9 + 2 = 11; so g = 1 + c = 1 + 11 = 12. Define total number of adult animals in Jefferson Circus as P; so P = g = 12. Define adult wolf in Mayer Aquarium as I; so I = P = 12. Define adult racoon in Mayer Aquarium as w; so w = g = 12. Define total number of adult animals in Mayer Aquarium as e; so e = I + w = 12 + 12 = 24. Answer: 24.