Question

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

$\frac{2}{x+2}+\frac{3}{x-1}=-\frac{8}{5}$

Short answer

The required solutions are $x=-4,\hspace{0.17em}x=-\frac{1}{8}$and the graph of the function is shown below:

Step-by-step solution

Step 1. Given information.

We have:

$\frac{2}{x+2}+\frac{3}{x-1}=-\frac{8}{5}$

Step 2. Simplify the equation.     

The equation is $\frac{2}{x+2}+\frac{3}{x-1}=-\frac{8}{5}$.

Multiply both sides by LCM (least common multiple): $\left(x+2\right)\left(x-1\right)$

$$\begin{gathered} \left(x+2\right)\left(x-1\right)\frac{2}{x+2}+\left(x+2\right)\left(x-1\right)\frac{3}{x-1}=\left(x+2\right)\left(x-1\right)-\frac{8}{5} \\ 10\left(x-1\right)+15\left(x+2\right)=-8\left(x+2\right)\left(x-1\right) \\ 10x-10+15x+30=-8\left(x^2+x-2\right) \\ 25x+20=-8x^2-8x+16 \\ 25x+20+8x^2+8x-16=0 \\ 8x^2+33x+4=0 \end{gathered}$$

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 $a=8,\hspace{0.17em}b=33,\hspace{0.17em}c=4$

$$\begin{gathered} x_{1,\hspace{0.17em}2}=\frac{-33\pm \sqrt{{33}^2-4·\hspace{0.17em}8·\hspace{0.17em}4}}{2·\hspace{0.17em}8} \\ {x}_{1,\hspace{0.17em}2}=\frac{-33\pm \hspace{0.17em}31}{2·\hspace{0.17em}8} \\ x=-\frac{1}{8},\hspace{0.17em}x=-4 \end{gathered}$$

Step 3. Verify the solution using a graphing utility.     

The solution of the equation is $x=-\frac{1}{8},\hspace{0.17em}x=-4$.

Draw the graph of the equation $\frac{2}{x+2}+\frac{3}{x-1}=-\frac{8}{5}$using graphing utility: