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# The perimeter of a rectangle is 42 inches. The length is 5 inches longer than the width. Find the length and width of the rectangle.

I need to know how to do this problem and how to set it up and solve it

thank you! :)

### 2 Answers by Expert Tutors

Francisco E. | Francisco; Civil Engineering, Math., Science, Spanish, Computers.Francisco; Civil Engineering, Math., Sci...
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The equations are:
2 x a +2 x b= 42
a = b+5
replacing a into the first equation we obtain
2(b + 5) + 2 x b= 42
4 x b = 32
b = 8
a = 13
Perimeter
26+16 = 42
CHECK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Micah D. | Dedicated Tutor for Reading, Writing, Math, and Test PrepDedicated Tutor for Reading, Writing, Ma...
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Since we are trying to find the values of two variables (the length L and the width W), we need at least two equations.  The perimeter of the rectangle is 42 inches, and the length is 5 inches longer than the width. Mathematically, we get our two equations from these two statements.

First, since the perimeter is 42 inches, we add the four sides and get L + L + W + W = 42.  Adding like terms gives us

2L + 2W = 42

Next, the length is 5 inches longer than the width, so

L - 5 = W

We then solve these two equations for L and W.  I will use substitution, but elimination would also work.

2L + 2W = 42
L - 5 = W

2L + 2(L-5) = 42
2L + 2L - 10 = 42
4L - 10 = 42
4L = 52
L = 13

Having solved for L, we can now substitute 13 for L in L - 5 = W to solve for W.

13 - 5 = W
W = 8

Our final step is to check by plugging our values into the original equations and making sure they work:

2(13) + 2(8) = 42   (Correct)
13 - 5 = 8   (Correct)