Chapter 7.5, Problem 10E

To determine

To find: The solution of the equation by cross multiplying.

Answer to Problem 10E

The solution of the given equation are $-1.5\&5$ .

Explanation of Solution

Given:

The given equation is:

  $\frac{-1}{x-3}=\frac{x-4}{x^2-27}$

The solution is calculated as:

  $\begin{array}{c}\frac{-1}{x-3}=\frac{x-4}{x^2-27}\\ -1\left(x^2-27\right)=\left(x-3\right)\left(x-4\right)\left[\text{cross multiply}\right]\\ -x^2+27=x^2-3x-4x+12\\ 2x^2-7x-15=0\end{array}$

Further simplified as:

  $\begin{array}{c}2x^2-10x+3x-15=0\\ \left(2x+3\right)\left(x-5\right)=0\\ \left(2x+3\right)=0\text{or}\\ \left(x-5\right)=0\end{array}$

Further simplified as:

  $\begin{array}{c}x=5\text{or}\\ x=-\frac{3}{2}\text{or}-\text{1}\text{.5}\end{array}$

Check for $x=-1.5$ is the solution of the equation.

  $\begin{array}{c}\frac{-1}{x-3}=\frac{x-4}{x^2-27}\\ \frac{-1}{-1.5-3}=\frac{-1.5-4}{{\left(-1.5\right)}^2-27}\\ 0.22=0.22\left[\text{true}\right]\end{array}$

Check for $x=5$ is the solution of the equation.

  $\begin{array}{c}\frac{-1}{x-3}=\frac{x-4}{x^2-27}\\ \frac{-1}{5-3}=\frac{5-4}{5^2-27}\\ -\frac{1}{2}=-\frac{1}{2}\left[\text{true}\right]\end{array}$

Hence, the solution of the given equation are $-1.5\&5$ .