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Lubov Fominskaja [6]
3 years ago
9

Solve this problem.

Mathematics
1 answer:
katrin [286]3 years ago
5 0
Let x = the width of the smaller rectangle. 
<span>The length of the smaller rectangle is 2x - 1. </span>
<span>Area is A = lw </span>
<span>So the area of the smaller rectangle is A = (x)(2x - 1) = 2x^2 - x </span>

<span>The larger rectangle's width is two inches more than the width of the smaller rectangle (x+2). </span>
<span>The larger rectangle's length is two inches more than the length of the smaller rectangle: </span>
<span>2x - 1 + 2 = 2x + 1 </span>
<span>Area is A = lw </span>
<span>The area of the larger rectangle is A = (x + 2)(2x + 1) = 2x^2 + x + 4x + 2 = 2x^2 + 5x + 2. </span>

<span>The area of the larger rectangle minus the area of the smaller rectangle is 86: </span>
<span>(2x^2 + 5x + 2) - (2x^2 - x) = 86 </span>

<span>Rewrite as adding the opposite: </span>
<span>(2x^2 + 5x + 2) + (-2x^2 + x) = 86 </span>

<span>Combine like terms: </span>
<span>6x + 2 = 86 </span>
<span>6x = 84 </span>
<span>x = 14 </span>

<span>The area of the smaller rectangle was 2x^2 - x, so </span>
<span>2(14)^2 - (14) </span>
<span>2(196) - 14 </span>
<span>392 - 14 </span>
<span>378 </span>

<span>The area of the smaller rectangle is 378 square inches.</span>
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