SOLUTION: The length of a rectangle is 4in longer than its width. If the perimeter of the rectangle is 28in, find its area.

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Question 217213: The length of a rectangle is 4in longer than its width. If the perimeter of the rectangle is 28in, find its area.
Found 2 solutions by rfer, ichudov:
Answer by rfer(16322) About Me  (Show Source):
You can put this solution on YOUR website!
2x+2(x+4)=28
2x+2x+8=28
2x=20
x=5
x+4=9
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A=lw
A=9*5
A=45 sq in

Answer by ichudov(507) About Me  (Show Source):
You can put this solution on YOUR website!
Keep in mind that perimeter is 2*length plus 2*width.
system%28+length+=+width%2B4%2C+2%2Alength%2B2%2Awidth+=+28+%29+
This is a linear system:
system%28+length+-width+=+4%2C+2%2Alength%2B2%2Awidth+=+28+%29+
which you can solve with the linear solver
Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
We'll use substitution. After moving -1*width to the right, we get:
1%2Alength+=+4+-+-1%2Awidth, or length+=+4%2F1+-+-1%2Awidth%2F1. Substitute that
into another equation:
2%2A%284%2F1+-+-1%2Awidth%2F1%29+%2B+2%5Cwidth+=+28 and simplify: So, we know that width=5. Since length+=+4%2F1+-+-1%2Awidth%2F1, length=9.

Answer: system%28+length=9%2C+width=5+%29.