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Let's call L the width of the rectangle and W its width. The area of the rectangle is the product between the length and the width, and we are also told that the area is 300 square meters, so we can write
Moreover, we know that the length is 5 meters longer than the width:
We have a system of 2 equations in 2 unknown variables, L and W. If we substitute the second equation into the first one, we get
which has two solutions: W=-20 and W=15. We can discard the negative solution since it does not have physical meaning, and now we can substitute the value of W into the second equation to find L:
<span>Therefore, the rectangle has width 15 meters and length 20 meters.</span>
Moreover, we know that the length is 5 meters longer than the width:
We have a system of 2 equations in 2 unknown variables, L and W. If we substitute the second equation into the first one, we get
which has two solutions: W=-20 and W=15. We can discard the negative solution since it does not have physical meaning, and now we can substitute the value of W into the second equation to find L:
<span>Therefore, the rectangle has width 15 meters and length 20 meters.</span>