Chapter 5, Problem 27REAE

Solution

To calculate: The solution of $\frac{x+12}{3}=\frac{2x+3}{x+2}$ by cross multiplication and check for extraneous solutions.

The required solution is $x=-5$ and $x=-3$ and it is not extraneous.

Given information:

The equation $\frac{x+12}{3}=\frac{2x+3}{x+2}$.

Calculation:

Consider the equation $\frac{x+12}{3}=\frac{2x+3}{x+2}$.

Cross multiply.

  $\left(x+12\right)\left(x+2\right)=3\left(2x+3\right)$

Use distributive property.

  $x^2+2x+12x+24=6x+9$

Simplify

  $\begin{array}{l}{x}^2+14x+24=6x+9\\ x^2+8x+15=0\\ x^2+5x+3x+15=0\\ \left(x+5\right)\left(x+3\right)=0\end{array}$

So, the solution is $x=-5$ or $x=-3$

To check, substitute $x=-5$ in $\frac{x+12}{3}=\frac{2x+3}{x+2}$.

  $\begin{array}{c}\frac{-5+12}{3}=\frac{2\left(-5\right)+3}{-5+2}\\ \frac{7}{3}=\frac{-10+3}{-3}\\ \frac{7}{3}=\frac{7}{3}\end{array}$

To check, substitute $x=-3$ in $\frac{x+12}{3}=\frac{2x+3}{x+2}$.

  $\begin{array}{c}\frac{-3+12}{3}=\frac{2\left(-3\right)+3}{-3+2}\\ \frac{9}{3}=\frac{-6+3}{-1}\\ 3=3\end{array}$

Here, the left hand side of the equation is equal to the right hand side and thus $x=-5$ and $x=-3$ are not extraneous solution.

Therefore, the required solution is $x=-5$ and $x=-3$ and it is not extraneous.