Chapter 7.5, Problem 27E

To determine

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

Answer to Problem 27E

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

Explanation of Solution

Given:

The equation is $\frac{x+1}{x+6}+\frac{1}{x}=\frac{2x+1}{x+6}$ .

Calculation:

In the equation $\frac{x+1}{x+6}+\frac{1}{x}=\frac{2x+1}{x+6}$ , the LCD is $x+6$ .

Multiply both sides by the LCD to remove the denominator.

  $\begin{array}{c}\frac{\left(x+1\right)\left(x+6\right)}{x+6}+\frac{x+6}{x}=\frac{\left(2x+1\right)\left(x+6\right)}{x+6}\\ x\left(x+1\right)+\left(x+6\right)=x\left(2x+1\right)\\ x^2+x+x+6=2x^2+x\\ x^2-x-6=0\end{array}$

Solve the equation $x^2-x-6=0$ for $x$ .

  $\begin{array}{c}{x}^2-x-6=0\\ x^2-3x+2x-6=0\\ \left(x-3\right)\left(x+2\right)=0\\ x=3,-2\end{array}$

Substitute 3 for $x$ in the equation $\frac{x+1}{x+6}+\frac{1}{x}=\frac{2x+1}{x+6}$ to verify the solution.

  $\begin{array}{c}\frac{3+1}{3+6}+\frac{1}{3}=\frac{2\left(3\right)+1}{3+6}\\ \frac{4}{9}+\frac{1}{3}=\frac{7}{9}\\ \frac{1}{3}=\frac{3}{9}\\ \frac{1}{3}=\frac{1}{3}\end{array}$

As LHS is equal to RHS so, $x=3$ is the solution of the given equation.

Substitute $-2$ for $x$ in the equation $\frac{x+1}{x+6}+\frac{1}{x}=\frac{2x+1}{x+6}$ to verify the solution.

  $\begin{array}{c}\frac{-2+1}{-2+6}+\frac{1}{-2}=\frac{2\left(-2\right)+1}{-2+6}\\ -\frac{1}{4}-\frac{1}{2}=-\frac{3}{4}\\ \frac{-1-2}{4}=-\frac{3}{4}\\ -\frac{3}{4}=-\frac{3}{4}\end{array}$

As LHS is equal to RHS so, $x=-2$ is the solution of the given equation.

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