Question

In Problem 17-36, solve each equation algebraically. Verify your solution using a graphing utility.

$\frac{1}{x+1}-\frac{5}{x-4}=\frac{21}{4}$

Short answer

The required solutions are $x=3,\hspace{0.17em}x=-\frac{16}{21}$. The graph of the solutions are given below:

Step-by-step solution

Step 1. Given information. 

We have:

$\frac{1}{x+1}-\frac{5}{x-4}=\frac{21}{4}$

Step 2. Simplify the equation.  

The equation is $\frac{1}{x+1}-\frac{5}{x-4}=\frac{21}{4}$.

Multiply both sides by LCM (least common multiple): $\left(x+1\right)\left(x-4\right)$

$$\begin{gathered} 4\left(x-4\right)-20\left(x+1\right)=21\left(x+1\right)\left(x-4\right) \\ 4x-16-20x-20=21x^2+21\left(-3x\right)+21\left(-4\right) \\ -16x-36=21x^2-63x-84 \\ 21x^2-63x-84+16x+36=0 \\ 21x^2-47x-48=0 \end{gathered}$$

Step 3. Solve the equation by quadratic formula. 

For a quadratic equation of the form $a{x}^2+bx+c=0$the solutions are:

$x_{1,\hspace{0.17em}2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

For $a=21,\hspace{0.17em}b=-47,\hspace{0.17em}c=-48$

$$\begin{gathered} x_{1,\hspace{0.17em}2}=\frac{-\left(-47\right)\pm \sqrt{{\left(-47\right)}^2-4·\hspace{0.17em}21\left(-48\right)}}{2·\hspace{0.17em}21} \\ {x}_{1,\hspace{0.17em}2}=\frac{-\left(-47\right)\pm \hspace{0.17em}79}{2·\hspace{0.17em}21} \\ x=3,\hspace{0.17em}x=-\frac{16}{21} \end{gathered}$$

Step 3. Verify the solution using a graphing utility. 

The solutions of the equation are $x=3,\hspace{0.17em}x=-\frac{16}{21}$.

Draw the graph of the equation $\frac{1}{x+1}-\frac{5}{x-4}=\frac{21}{4}$using graphing utility: