Q. 461 - Intermediate Algebra

Question

The area of a rectangular placemat is $168$ square inches. Its length is two inches longer than the width. Find the length and width of the placemat.

Step-by-Step Solution

Answer

The length and width of the placemat is $14$ inches and $12$ inches.

Step 1. Assume the integers.

Assume width of the placemat to be $x$ inches and length of the placemat to be $x+2$ inches.

Area of placemat$=length\times width$

Substitute the values in the above equation,

$$\begin{gathered} 168=x\left(x+2\right) ...... \left(i\right) \\ 168=x^2+2x \end{gathered}$$

Bring all the terms to one side,

$x^2+2x-168=0$

Step 2. Factor the equation.

On factoring,

$$\begin{gathered} x^2+14x-12x-168=0 \\ x\left(x+14\right)-12\left(x+14\right)=0 \\ \left(x+14\right)\left(x-12\right)=0 \end{gathered}$$

Then,

$$\begin{gathered} x+14=0 \\ x=-14 \end{gathered}$$

As we know, width cannot be negative. Hence, it is not considered.

$$\begin{gathered} x-12=0 \\ x=12inches \end{gathered}$$

Then the length,

$$\begin{gathered} x+12=12+2 \\ =14inches \end{gathered}$$

Step 3. Check the answers.

Substitute the obtained values of the length and width in equation (i),

$$\begin{gathered} 14\times 12=168 \\ 168=168 \end{gathered}$$

This is true.