SOLUTION: Could you please help me solve this problem. I tried to solve it by using the given measurements to find the area and then plugging the variable into an equation that would work fo

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Question 179881This question is from textbook Merril Algebra 1
: Could you please help me solve this problem. I tried to solve it by using the given measurements to find the area and then plugging the variable into an equation that would work for the value of the variable. It did not work and left me just as confused. The problem states:"Mr. Herrera had a concrete sidewalk built on three sides of his yard as shown at the right. The yard measures 24 feet by 42 feet. The longer walk is 3 feet wide. The price of the concrete was $22 per square yard, and the total bill was $902. What is the width of the walk on the remaining two sides? This question is from textbook Merril Algebra 1

Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

This was coming out very ugly until I realized that the sidewalk must go around the outside of the 42ft by 24ft yard rather than being a part of that area.

Let x feet be the dimension of the side walkways. There are two of them so they add 2x feet to the overall dimension of 42 feet. So the new long dimension is . The single long walkway is 3 feet wide, so it adds 3 feet to the 24 foot dimension making 27 feet. That means the total area is given by .

Since he paid $902 for the concrete at $22/sq yard, he must have added to the area of the yard. There are 9 square feet in 1 square yard, so he added to the area of the yard. The original area was . 369 + 1008 = 1377.

So:



Solve for x:










John
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