Chapter 7.5, Problem 30E

To determine

To solve: The equation by using the LCD and check for solution.

Answer to Problem 30E

The solutions of the equation $\frac{5}{x^2+x-6}=2+\frac{x-3}{x-2}$ are $x=\frac{-1+\sqrt{79}}{3}$ and $x=\frac{-1-\sqrt{79}}{3}$ .

Explanation of Solution

Given:

The equation is $\frac{5}{x^2+x-6}=2+\frac{x-3}{x-2}$ .

Calculation:

Rewrite the given equation as $\frac{5}{\left(x+3\right)\left(x-2\right)}=2+\frac{x-3}{x-2}$

In the equation $\frac{5}{\left(x+3\right)\left(x-2\right)}=2+\frac{x-3}{x-2}$ , the LCD is $\left(x+3\right)\left(x-2\right)$ .

Multiply both sides by the LCD to remove the denominator.

  $\begin{array}{c}\frac{5\left(x+3\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}=2\left(x+3\right)\left(x-2\right)+\frac{\left(x-3\right)\left(x+3\right)\left(x-2\right)}{x-2}\\ 5=2\left(x+3\right)\left(x-2\right)+\left(x-3\right)\left(x+3\right)\\ 5=2x^2+2x-12+x^2-9\\ 3x^2+2x-26=0\end{array}$

Solve the equation $3x^2+2x-26=0$ by using Quadratic Formula for $x$ .

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-2\pm \sqrt{2^2-4\left(3\right)\left(-26\right)}}{2\left(3\right)}\\ =\frac{-2\pm \sqrt{316}}{6}\\ =\frac{-1+\sqrt{79}}{3},\frac{-1-\sqrt{79}}{3}\text{or}2.63,-3.3\end{array}$

Substitute $2.63$ for $x$ in the equation

  $\frac{5}{x^2+x-6}=2+\frac{x-3}{x-2}$ to verify the solution.

  $\begin{array}{c}\frac{5}{{\left(2.63\right)}^2+\left(2.63\right)-6}=2+\frac{2.63-3}{2.63-2}\\ 1.41=1.41\end{array}$

As LHS is equal to RHS so,

  $x=\frac{-1+\sqrt{79}}{3}$ is the solution of the given equation.

Substitute $-3.3$ for $x$ in the equation

  $\frac{5}{x^2+x-6}=2+\frac{x-3}{x-2}$ to verify the solution.

  $\begin{array}{c}\frac{5}{{\left(-3.3\right)}^2-3.3-6}=2+\frac{-3.3-3}{-3.3-2}\\ 3.1\approx 3.1\end{array}$

As LHS is equal to RHS so, $x=\frac{-1-\sqrt{79}}{3}$ is the solution of the given equation.

Therefore, the solutions of the equation $\frac{5}{x^2+x-6}=2+\frac{x-3}{x-2}$ are $x=\frac{-1+\sqrt{79}}{3}$ and $x=\frac{-1-\sqrt{79}}{3}$ .