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The length of the rectangle is $4$ m longer than its width. If the area is $8~\text{m}^2$, find the rectangle's dimensions. Round to the nearest $10$th of a meter.

I have absolutely no clue how to begin with this.

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    $\begingroup$ $L=W+4$ and $8=LW=(W+4)W=W^2+4W$. So, $W=?$ and $L=W+4=?$ $\endgroup$ Jan 26, 2016 at 2:30

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Well, for word problems, a good idea to begin with is translating the English sentences into math, in this case, equations.

The length of the rectangle is 4 longer than its width

$L = W + 4$ where $L$ and $W$ denote the length and width.

The area is 8

$A = 8 = LW$ where the latter equation is the formula for area.

So we know that $L = W+4$ which allows us to plug it into the area equation and solve for $W$.

$$8 = (W+4)W$$

and once you have $W$, you can plug it back into $L= W+4$ to solve for $L$.

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