Chapter 7.5, Problem 28E

To determine

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

Answer to Problem 28E

The solutions of the equation $\frac{x+3}{x-3}+\frac{x}{x-5}=\frac{x+5}{x-5}$ are $x=0$ and $x=7$ .

Explanation of Solution

Given:

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

Calculation:

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

Multiply both sides by the LCD to remove the denominator.

  $\begin{array}{c}\frac{\left(x+3\right)\left(x-3\right)\left(x-5\right)}{x-3}+\frac{x\left(x-3\right)\left(x-5\right)}{x-5}=\frac{\left(x+5\right)\left(x-3\right)\left(x-5\right)}{x-5}\\ \left(x+3\right)\left(x-5\right)+x\left(x-3\right)=\left(x+5\right)\left(x-3\right)\\ x^2-2x-15+x^2-3x=x^2+2x-15\\ x^2-8x=0\end{array}$

Solve the equation $x^2-8x=0$ for $x$ .

  $\begin{array}{c}{x}^2-7x=0\\ x\left(x-7\right)=0\\ x=0,7\end{array}$

Substitute 0 for $x$ in the equation $\frac{x+3}{x-3}+\frac{x}{x-5}=\frac{x+5}{x-5}$ to verify the solution.

  $\begin{array}{c}\frac{0+3}{0-3}+\frac{0}{x-5}=\frac{0+5}{0-5}\\ -1=-1\\ 1=1\end{array}$

As LHS is equal to RHS so, $x=0$ is the solution of the given equation.

Substitute 7 for $x$ in the equation $\frac{x+3}{x-3}+\frac{x}{x-5}=\frac{x+5}{x-5}$ to verify the solution.

  $\begin{array}{c}\frac{7+3}{7-3}+\frac{7}{7-5}=\frac{7+5}{7-5}\\ \frac{10}{4}+\frac{7}{2}=\frac{12}{2}\\ \frac{12}{2}=\frac{12}{2}\\ 6=6\end{array}$

As LHS is equal to RHS so, $x=7$ is the solution of the given equation.

Therefore, the solutions of the equation $\frac{x+3}{x-3}+\frac{x}{x-5}=\frac{x+5}{x-5}$ are $x=0$ and $x=7$ .