Question

A rectangle's length, in inches, is 6
more than twice its width, in inches. If the perimeter of the rectangle is 60
inches, what is its width?

Answer 1

Let the width of the rectangle be denoted as \( w \) inches. According to the problem, the length \( l \) of the rectangle can be expressed as:

\[
l = 2w + 6
\]

The formula for the perimeter \( P \) of a rectangle is given by:

\[
P = 2(l + w)
\]

We know the perimeter is 60 inches, so we set up the equation:

\[
2(l + w) = 60
\]

Dividing both sides by 2, we get:

\[
l + w = 30
\]

Now we can substitute the expression for \( l \) from the first equation:

\[
(2w + 6) + w = 30
\]

Combining like terms, we have:

\[
3w + 6 = 30
\]

Next, we subtract 6 from both sides:

\[
3w = 24
\]

Now, we can divide both sides by 3 to solve for \( w \):

\[
w = 8
\]

Thus, the width of the rectangle is \( \boxed{8} \) inches.