SOLUTION: Could you please solve this equation for me. Suppose that the width if a rectangle is 5 inches shorter than the length and tha perimeter of the rectangle is 50. What is the w

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Question 119916This question is from textbook Blitzer College Algebra
: Could you please solve this equation for me.
Suppose that the width if a rectangle is 5 inches shorter than the length and tha perimeter of the rectangle is 50. What is the width? Could you please show you work so I can get a better understanding? Thanks This question is from textbook Blitzer College Algebra

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You can start with the formula for the perimeter of a rectangle:
P+=+2%28L%2BW%29 where L = length and W = width.
The problem states that the width, W, is 5 inches shorter than the length, L.
You write this as:
W+=+L-5 but since we are looking for the width, W, we'll rewrite this as:
L+=+W%2B5
The perimeter, P, is given as 50, so P+=+50
Now we'll substitute these values into the formula above and solve for W.
P+=+2%28L%2BW%29 Substitute L+=+W%2B5
P+=+2%28%28W%2B5%29%2BW%29 Combine like-terms.
P+=+2%282W%2B5%29 Apply the distributive property.
P+=+4W%2B10 Substitute P = 50.
50+=+4W%2B10 Subtract 10 from both sides.
40+=+4W Finally, divide both sides by 4.
10=+W
The width is 10 inches.
L+=+W%2B5
L+=+10%2B5
L+=+15
The length is 15 inches.
Check:
P+=+2%28L%2BW%29 Substitute L= 15 and W = 10
P+=+2%2815%2B10%29
P+=+2%2825%29
P+=+50 The solution is good!