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.


`(8w)/(8)= (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