SOLUTION: A frame around a family portrait has a perimeter of 98 inches. The length of the frame is 8 inches less than twice the width. Find the length and width of the frame.

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Question 1151916: A frame around a family portrait has a perimeter of 98 inches. The length of the frame is 8 inches less than twice the width. Find the length and width of the frame.
Found 3 solutions by josgarithmetic, MathLover1, MathTherapy:
Answer by josgarithmetic(39630) About Me  (Show Source):
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

A frame around a family portrait has a perimeter of 98 inches.
2%28L%2BW%29=98
L%2BW=98%2F2
L%2BW=49
L=49-W.......eq.1
if the length of the frame is+8+inches+less than twice the width, we have
L=2W-8.......eq.2
from eq.1 and eq.2 we have
49-W=2W-8........solve for W
49%2B8=2W%2BW
57=3W
W=57%2F3
W=19
then
L=49-19.......eq.1
L=30
the length of the frame is 30 and width of the frame is 19

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!

A frame around a family portrait has a perimeter of 98 inches. The length of the frame is 8 inches less than twice the width. Find the length and width of the frame.
Let width be W

Then length is: 2W - 8

With perimeter being 98, length + width = matrix%281%2C3%2C+98%2F2%2C+%22=%22%2C+49%29

We then get: 2W  - 8 + W = 49

3W = 57

W, or