Question

The length of a rectangle is 8 inches shorter than three times its width. If the area of the rectangle is 35 square inches. Find the dimensions of the rectangle

Answer 1

The dimensions of the rectangle are 7 inches by 5 inches;

`\begin{gathered} \text{length = 7 inches} \\ \text{width = 5 inches} \end{gathered}`

Step-by-step explanation:

Given that the length of a rectangle is 8 inches shorter than three times its width.

Let x represent its length and w represent its width.

`x=3w-8\text{ -----1}`

If the area of the rectangle is 35 square inches;

`xw=35\text{ ----2}`

Making w the subject of formula in equation 2;

`w=(35)/(x)`

substituting into equation 1;

`\begin{gathered} x=3w-8\text{ -----1} \\ x=3((35)/(x))-8 \\ \text{multiply through by x;} \\ x^2=105-8x \\ x^2+8x-105=0 \end{gathered}`

Solving for x in the quadratic equation;

`\begin{gathered} (x+15)(x-7)=0 \\ x=-15 \\ \text{and} \\ x=7 \end{gathered}`

Since length cannot be negative;

`x=7`

substituting to equation 2;

`\begin{gathered} xw=35 \\ 7w=35 \\ w=(35)/(7) \\ w=5 \end{gathered}`

Therefore, the dimensions of the rectangle are 7 inches by 5 inches;

`\begin{gathered} \text{length = 7 inches} \\ \text{width = 5 inches} \end{gathered}`