Question

A rectangle has a length one less than three times the width. If the perimeter is 46 units , find the width and length

Answer 1

Greetings!

Let l represent the length of the rectangle.
Let w represent the width of the rectangle.

$\left \{ {{2(l+w)=46} \atop {3w-1=l}} \right.$

Isolate for the l in Equation #2.
$3w-1=l$
$l=3w-1$

Substitute Equation #2 into Equation #1.
$2(l+w)=46$

$2[(3w-1)+w)]=46$

Combine Like Terms.
$2(4w-1)=46$

Distribute the Parenthesis.
How? Multiply the terms inside the Parenthesis by the term outside of the Parenthesis.
$2*4w-2*1=46$

$8w-2=46$

Add 2 to both sides.
$(8w-2)+2=(46)+2$

$8w=48$

Divide both sides by 8.

$\frac{8w}{8}= \frac{48}{8}$

$w=6$

The Answer Is:
$\left[\begin{array}{ccc}w=6\end{array}\right]$

The Width is 6 units and the Length is 17 units.

Hope this helps.
-Benjamin