Let's call x the width of the rectangle and y the length of the rectangle.
The width of the rectangle is 12 less than the length, so:
x = y - 12 equation 1
Also, the perimeter is 156 inches, so:
2x + 2y = 156 equation 2
Replacing equation 1 on equation 2, and solving for y, we get:
[tex]\begin{gathered} 2(y-12)+2y=156 \\ 2y-24+2y=156 \\ 4y\text{ -24=156} \end{gathered}[/tex][tex]\begin{gathered} 4y-24+24=156+24 \\ 4y=180 \\ \frac{4y}{4}=\frac{180}{4} \\ y=45 \end{gathered}[/tex]
Then, replacing the value of y on the first equation, we get:
x = 45 - 12
x = 33
Finally, the width is 33 and the length is 45 inches.
