```Question 171289
width of rectangle equals 15 centimeters.
length of rectangle equals 18 centimeters.
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let L = length
let W = width
let A = area.
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area of triangle = length times width:
A = L * W
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If both dimensions are decreased by the same amount, the area of the new rectangle is 116cm squared less than the area of the original.
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let x be the amount the length and the width are decreased by.
your formula to solve would be:
A - 116 = (L-x) * (W-x)
since A is the original area and it is equal to L * W, you can substitute in the equation to get:
(L*W) - 116 = (L-x)*(W-x)
since you know what L is and you know what W is you can substitute for them in the equation:
(18*15) - 116 = (18-x)*(15-x)
simplify:
270 - 116 = 270 - 18*x - 15*x + x^2
subtract 270 from both sides and it simplify further:
- 116 = -33*x + x^2
add 116 to both sides and rearrange to the standard order exponents (highest order first):
x^2 - 33*x + 116 = 0
x^2 - 33*x + 116 = 0
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i didn't see it right away, so i used the quadratic formula of:
{{{x = ((-b)+-sqrt(b^2-4ac))/(2a)}}}
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the standard form of the quadratic equation is:
{{{a*x^2 + b*x + c = 0}}}
a is the coefficient of x^2
b is the coefficient of x
c is the constant term
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a = 1
b = -33
c = 116
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the factors came out to be:
(x-29) * (x-4) = 0
x = 29
or:
x = 4
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to see which of these answers is correct, we need to substitute them in the original equation.
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the original equation is:
(18*15) - 116 = (18-x)*(15-x)
substitute 29 for x:
(18*15) - 116 = (18-29)*(15-29)
simplify:
270 - 116 = (-11)*(-14)
154 = 154
x = 29 works.
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substitute 4 for x:
(18*15) - 116 = (18-4)*(15-4)
simplify:
270 - 116 = (14)*(11)
154 = 154
x = 4 works as well.
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