Question

The length of a rectangle is five more than its width. if the perimeter is 120, find both the length and the width.

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Neal

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Let's represent the width of the rectangle as $w$.

According to the problem, the length is five more than the width, so we can write the equation: length = $w+5$.

The perimeter of a rectangle is equal to twice the sum of its length and width. So, the equation for the perimeter is: $2(w + (w+5)) = 120$.

Simplifying the equation:

$2(2w + 5) = 120 \implies 4w + 10 = 120$.

Subtracting 10 from both sides:

$4w = 110$.

Dividing by 4 on both sides:

$w = 27.5$.

Thus, the width of the rectangle is 27.5.

To find the length, we substitute the value of $w$ back into the equation:

length = $w + 5 = 27.5 + 5 = 32.5$.

Answer: The width of the rectangle is 27.5 units, and the length is 32.5 units.

According to the problem, the length is five more than the width, so we can write the equation: length = $w+5$.

The perimeter of a rectangle is equal to twice the sum of its length and width. So, the equation for the perimeter is: $2(w + (w+5)) = 120$.

Simplifying the equation:

$2(2w + 5) = 120 \implies 4w + 10 = 120$.

Subtracting 10 from both sides:

$4w = 110$.

Dividing by 4 on both sides:

$w = 27.5$.

Thus, the width of the rectangle is 27.5.

To find the length, we substitute the value of $w$ back into the equation:

length = $w + 5 = 27.5 + 5 = 32.5$.

Answer: The width of the rectangle is 27.5 units, and the length is 32.5 units.

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