SOLUTION: Suppose that the width of a rectangle is 5 inches shorter than the length and that the perimeter of the rectangle is 50. Part A: Setup an equationfor the perimeter involving o

Question 88975: Suppose that the width of a rectangle is 5 inches shorter than the length and that the perimeter of the rectangle is 50.
Part A: Setup an equationfor the perimeter involving only L, the length of the rectangle.
Part B: Solve this equation algebraically to find the length of the rectangle. Find the width as well.
Answer by checkley75(3666) (Show Source): You can put this solution on YOUR website!
W=L-5
2W+2L=50
2(L-5)+2L=50
2L-10+2L=50
4L-10=50
4L=50+10
4L=60
L=60/4
L=15 ANSWER FOR THE LENGTH.
W=15-5
W=10 ANSWER FOR THE WIDTH.
PROOF
2*10+2*15=50
20+30=50
50=50

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