SOLUTION: Find the length a d the width of the rectangle described. The length is 12 units more than the width. The perimeter is 7 times the width.

Algebra ->  Test -> SOLUTION: Find the length a d the width of the rectangle described. The length is 12 units more than the width. The perimeter is 7 times the width.      Log On


   



Question 491749: Find the length a d the width of the rectangle described.
The length is 12 units more than the width. The perimeter is 7 times the width.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

The length L is 12 units more than the width W.
L=W%2B12.......1
The perimeter P is 7 times the width W.
P=7W
------------------------
P=2%28L%2BW%29
2%28L%2BW%29=7W...replace L with W%2B12
2%28W%2B12%2BW%29=7W
2%282W%2B12%29=7W
4W%2B24=7W
24=7W-4W
24=3W
24%2F3=W
highlight%288=W%29
now find L
L=W%2B12.......1
L=8%2B12
highlight%28L=20%29
check:
P=2%28L%2BW%29
P=2%2820%2B8%29
P=2%2828%29
P=56
and another one
P=7W
P=7%2A8
P=56