Chapter 5, Problem 28REAE

Solution

To calculate: The solution of $\frac{2x}{x+4}=\frac{-3x}{4x-3}$ by cross multiplication and check for extraneous solutions.

The required solution is $x=\frac{-6}{11}$ and it is not extraneous.

Given information:

The equation $\frac{2x}{x+4}=\frac{-3x}{4x-3}$.

Calculation:

Consider the equation $\frac{2x}{x+4}=\frac{-3x}{4x-3}$.

Cross multiply.

  $2x\left(4x-3\right)=-3x\left(x+4\right)$

Use distributive property.

  $8x^2-6x=-3x^2-12x$

Simplify

  $\begin{array}{c}11x^2+6x=0\\ 11x+6=0\\ 11x=-6\\ x=\frac{-6}{11}\end{array}$

So, the solution is $x=\frac{-6}{11}$.

To check, substitute $x=\frac{-6}{11}$ in $\frac{2x}{x+4}=\frac{-3x}{4x-3}$.

  $\begin{array}{c}\frac{2\left(\frac{-6}{11}\right)}{\frac{-6}{11}+4}=\frac{-3\left(\frac{-6}{11}\right)}{4\left(\frac{-6}{11}\right)-3}\\ \frac{\frac{-12}{11}}{\frac{-6+44}{11}}=\frac{\frac{18}{11}}{\frac{-24-33}{11}}\\ \frac{-12}{38}=\frac{18}{-57}\\ -\frac{6}{19}=-\frac{6}{19}\end{array}$

Here, the left hand side of the equation is equal to the right hand side and thus $x=\frac{-6}{11}$ is not extraneous solution.

Therefore, the required solution is $x=\frac{-6}{11}$ and it is not extraneous.