SOLUTION: I think this is right for this problem. Can someone pls check? 50/4=12.5x2+7=32 32/2=16-7=9 W=32 L=9 The length of a rectangle is 7" greater than its width. The perimete

Question 408536: I think this is right for this problem. Can someone pls check?
50/4=12.5x2+7=32
32/2=16-7=9
W=32
L=9
The length of a rectangle is 7" greater than its width. The perimeter of the rectangle is 50". Find the length and width of the rectangle.

Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The length of a rectangle is 7" greater than its width. The perimeter of the rectangle is 50". Find the length and width of the rectangle.
------
Perimeter = 2(length + width)
---
Let width be "x".
The length is "x+7".
---
50 = 2(x+7 + x)
25 = 2x+7
2x = 18
x = 9 (width)
x+7 = 16 (length)
======================
Cheers,
Stan H.
Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!

The length of a rectangle is greater than its width . The perimeter of the rectangle is .
Find the length and width of the rectangle.

given: of a rectangle is greater than its width ..=>

.....
.divide by 2

.....

.....
.....

check:

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