The answer is to the question is C
Let the width be x.
Length is 8 feet more than width. Length = x + 8
Area = x(x + 8)
width increased by 4, that is, (x + 4)
Length decreased by 5, (x + 8 - 5) = (x + 3)
Area = (x + 4)(x +3)
Area remains the same
x(x + 8) = (x+4)(x +3)
x² + 8x = x(x +3) + 4(x +3)
x² + 8x = x² +3x + 4x +12
x² + 8x = x² +7x +12 Eliminate x² from both sides
8x = 7x + 12
8x - 7x = 12
x = 12
Dimensions of original rectangle : x, x + 8
12, 12 +8 = 12, 20
Original rectangle is 20 feet by 12 feet
A + c = 824 represents number of tickets to sell
20a + 12c = 13,344 represents cost of each type of ticket
multiply equation 1 by -12 to cancel the variable 'c' and solve for 'a'
-12(a + c = 824)
-12a - 12c = - 9888
20a + 12c = 13,344 add equations together
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8a = 3456 divide both sides by 8
a = 432
To find children's tickets substitute adult numbers back into equation 1.
a + c = 824
432 + c = 824
subtract 432 from both sides
c = 392
adult 432, children 392
