Chapter 5.9, Problem 15WE

To determine

To solve: the given equation for $\mathrm{x}$ . Check the result.

Answer to Problem 15WE

The value of $\mathrm{x}$ in the given equation is $4$ .

Explanation of Solution

Given information:

Given equation is

  $\frac{3}{\mathrm{x}+1}-\frac{1}{\mathrm{x}-2}=\frac{1}{{\mathrm{x}}^2-\mathrm{x}-2}$

Calculation:

Consider the given equation,

  $\frac{3}{\mathrm{x}+1}-\frac{1}{\mathrm{x}-2}=\frac{1}{{\mathrm{x}}^2-\mathrm{x}-2}$

Now, solve the equation as shown,

  $\begin{array}{c}\frac{3}{\mathrm{x}+1}-\frac{1}{\mathrm{x}-2}=\frac{1}{{\mathrm{x}}^2-\mathrm{x}-2}\\ \frac{3\mathrm{x}-6-\mathrm{x}-1}{\left(\mathrm{x}+1\right)\left(\mathrm{x}-2\right)}=\frac{1}{{\mathrm{x}}^2-\mathrm{x}-2}\\ \frac{2\mathrm{x}-7}{{\mathrm{x}}^2-\mathrm{x}-2}=\frac{1}{{\mathrm{x}}^2-\mathrm{x}-2}\\ \left({\mathrm{x}}^2-\mathrm{x}-2\right)\left(2\mathrm{x}-7\right)={\mathrm{x}}^2-\mathrm{x}-2\\ \left({\mathrm{x}}^2-\mathrm{x}-2\right)\left(2\mathrm{x}-8\right)=0\\ 2\mathrm{x}=8\\ \mathrm{x}=4\end{array}$

Now, check the result by back substitution.

  $\begin{array}{l}\frac{3}{\mathrm{x}+1}-\frac{1}{\mathrm{x}-2}=\frac{1}{{\mathrm{x}}^2-\mathrm{x}-2}\\ \frac{3}{4+1}-\frac{1}{4-2}=\frac{1}{4^2-4-2}\\ \frac{3}{5}-\frac{1}{2}=\frac{1}{10}\\ \frac{6-5}{10}=\frac{1}{10}\\ \frac{1}{10}=\frac{1}{10}\end{array}$

Which is true.

Hence, the value of $\mathrm{x}$ in the given equation is $4$ .