Q. 327 - Intermediate Algebra

Question

The area of a bulletin board is $55$ feet. The length is four feet less than three times the width. Find the length and width of the bulletin board.

Step-by-Step Solution

Answer

The length and width of the bulletin board is $5 feet$ and $11 feet$.

Step 1. Assumption of integers.

Assume the width and the length of the board to be $x feet , 3x-4 feet$.

Area of bulletin board$=length\times width$

Substitute the values in the equation of area,

$$\begin{gathered} 55=x\left(3x-4\right) \\ 3x^2-4x=55 \\ 3x^2-4x-55=0 ...... \left(i\right) \end{gathered}$$

Step 2. Factor the equation.

On factoring,

$$\begin{gathered} 3x^2-15x+11x-55=0 \\ 3x\left(x-5\right)+11\left(x-5\right)=0 \\ \left(x-5\right)\left(3x+11\right)=0 \end{gathered}$$

Use the zero product property,

$$\begin{gathered} x-5=0 \\ x=5 \end{gathered}$$

Then,

$$\begin{gathered} 3x+11=0 \\ x=-\frac{11}{3} \end{gathered}$$

As we know, the width of the board cannot be negative, so $x=5$.

Then, the length of the board,

$$\begin{gathered} =3x-4 \\ =3\left(5\right)-4 \\ =15-4 \\ =11 \end{gathered}$$

Step 3. Check the answers.

Substitute the width and the length in equation (i),

$$\begin{gathered} 55=5\times 11 \\ 55=55 \end{gathered}$$

This is true.

Hence, the width and the length of the board is $5 feet, 11 feet$.