SOLUTION: the width of a rectangle is 40 cm less than its perimeter. The rectangle's are is 102 square cm. find the dementions

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Question 946751: the width of a rectangle is 40 cm less than its perimeter. The rectangle's are is 102 square cm. find the dementions
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
ASSIGN VARIABLES
w, width of rectangle
L, length of rectangle
p, perimeter of rectangle
A, area of rectangle
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A=102
w=p-40
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UNKNOWNS: w and L.

system%28w=p-40%2C2w%2B2L=p%2CwL=A%29

Note that although p is unknown, it can be eliminated using the w equation
and the p equation.

w=p-40
w%2B40=p
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Equate the two expressions for p;
w%2B40=2w%2B2L
Simplify
40=w%2B2L
w%2B2L=40

The simpler equivalent system without p is
system%28w%2B2L=40%2CwL=A%29

Substitute for w in the area A equation:
%2840-2L%29L=A
simplify
40L-2L%5E2-A=0
-40L%2B2L%5E2%2BA=0
highlight_green%282L%5E2-40L%2BA=0%29

Substituting the given value now for A would be convenient.
2L%5E2-40L%2B102=0
L%5E2-20L%2B51=0
FACTORABLE.
%28L-3%29%28L-17%29=0
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EITHER highlight%28L=3%29 OR highlight%28L=17%29.

Find width, w, using the earlier found formula from the process,
w=40-2L:
EITHER highlight%28w=34%29 OR highlight%28w=6%29


SUMMARY OF SOLUTIONS
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The dimensions are either 3 & 34, or 17 & 6.
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