L = W + 9
2L + 2W = 2(W+9) + 2W = 50
Solve for W, then L.
Solve each problem algebraically using two variables.
The perimeter of a rectangle is 50 centimeters. The length is 9 centimeters more than the width. Find the length and the width of the rectangle
2 answers
P= 50
L= W + 9
P=2L + 2W
P= 2*(W +9) +2W
50 = 2W+18+2W
4W = 50-18
4W = 32
W = 8
L = W+9
L = 8 + 9
L = 17
therefore, L= 17 and w =8
not sure if my work is correct
L= W + 9
P=2L + 2W
P= 2*(W +9) +2W
50 = 2W+18+2W
4W = 50-18
4W = 32
W = 8
L = W+9
L = 8 + 9
L = 17
therefore, L= 17 and w =8
not sure if my work is correct