SOLUTION: Suppose that the width of a rectangle is 2 inches shorter than the length and that the perimeter of the rectangle is 80.
a) Set up an equation for the perimeter involving only L,
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a) Set up an equation for the perimeter involving only L,
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Question 149942This question is from textbook
: Suppose that the width of a rectangle is 2 inches shorter than the length and that the perimeter of the rectangle is 80.
a) Set up an equation for the perimeter involving only L, the length of the rectangle.
b) Solve this equation algebraically to find the length of the rectangle. Find the width as well. This question is from textbook
You can put this solution on YOUR website! Suppose that the width of a rectangle is 2 inches shorter than the length and that the perimeter of the rectangle is 80.
a) Set up an equation for the perimeter involving only L, the length of the rectangle.
b) Solve this equation algebraically to find the length of the rectangle. Find the width as well.
.
Let L = length of rectangle
L-2 = width of rectangle
.
Since the perimeter is twice the "length + width" we have:
2(L + L - 2) = 80 (solution for a)
.
solving for L:
2(L + L - 2) = 80
(2L - 2) = 40
2(L - 1) = 40
L - 1 = 20
L = 21 inches (solution for b -- length of rectangle)
L-2 = 19 inches (solution for b -- width of rectangle)
