Question

The width of a rectangle is 7 inches less than its length. The area of the rectangle is 120 square inches. Solve for the dimensions of the rectangle.

Answer 1

\[120=(L)(L-7). Solve for L\]

Answer 2

\[L ^{2} - 7L= 120\] ?

Answer 3

Yes, that will give you the length \(L\). Remember length cannot be negative, so you should be left with a single solution. Once you've found it you can find the width because you're told \(W = L-7\)

Answer 4

what would the step after that be? adding 7L or... ?

Answer 5

You are asked to find the dimension of the rectangle which are the length \(L\) and width \(W\).
To do so you're told two things: The area is 120, and the width is 7 less than the length.
That means we have the following system of equations:
$$
\begin{cases}
L \cdot W = 120 \\
W = L - 7
\end{cases}
$$Now we need to solve it. We start by substituting \(W\) in the first equation using the second one \(W = L -7\) and we get:
$$
L \cdot W = 120 \implies L \cdot (L -7) = 120 \implies L^2 - 7L = 120 \\
L^2 -7L - 120 = 0
$$Now we have to solve for \(L\) just like any other quadratic equation, keeping in mind that \(L \ge 0\) because the length of a rectangle cannot be negative:
$$
L_{1,2} = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot (-120)}}{2} = \frac{7 \pm \sqrt{49 + 480}}{2} = \frac{7 \pm 23}{2} \\
L_1 = \frac{7 + 23}{2} = \frac{30}{2} = 15 \qquad \cancel{L_2 = \frac{7-23}{2} = \frac{-16}{2} = -8}
$$
Once we found \(L\) we know \(W\) because \(W = L-7\), so we can plug in:
$$W = L - 7 \implies W = 15 - 7 = 8$$
And that's it. The length is 15 and the width is 8, and we can verify that those values indeed match the properties of the rectangle: \(8 = 15 - 7 \qquad 15 \cdot 8 = 120\)

Answer 6

have u solved??