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Suppose that the width of a certain rectangle is 1 inch more than one-fourth of its length. The perimeter of the rectangle is 3...Asked by Anonymous
Suppose that the width of a certain rectangle is 1 inch more than one-fourth of its length. The perimeter of the rectangle is 62 inches. Find the length and width of the rectangle.
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Answered by
Steve
If the length is y, then the width w is 1 + y/4. So,
2(1 + y/4 + y) = 62
Now just solve for y, and you can find w.
2(1 + y/4 + y) = 62
Now just solve for y, and you can find w.
Answered by
Kassen
x=width, y=length => x-1=y*1/4; 2x+2y=62;
y=4x-4 => 2x+8x-8=62 => 10x=70 => x=7 (width) => y (length) = (62-14)/2 =24; let's check it: 7-1=24/4 <=> 6=6
y=4x-4 => 2x+8x-8=62 => 10x=70 => x=7 (width) => y (length) = (62-14)/2 =24; let's check it: 7-1=24/4 <=> 6=6