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marinap97
PROBLEM SOLVING USING SYSTEMS OF EQUATIONS The length of a rectangle is 5cm more than twice the width. The perimeter of the rectangle is 34 cm. Find the length and the width of the rectangle.
L=5+2w 2L+2w=34 substitute for L
let L = length of rectangle W = width of rectangle P = perimeter L = 5 + 2w 2L + 2W = 34 L - 2W = 5 2L +2W = 34 3L = 39 L = 13 W = 4
how would you do that with x and y?
just use x and y instead of L and W