Chapter 5.9, Problem 14WE

To determine

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

Answer to Problem 14WE

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

Explanation of Solution

Given information:

Given equation is

  $\frac{\mathrm{k}}{\mathrm{k}+1}+\frac{\mathrm{k}}{\mathrm{k}-2}=2$

Calculation:

Consider the given equation,

  $\frac{\mathrm{k}}{\mathrm{k}+1}+\frac{\mathrm{k}}{\mathrm{k}-2}=2$

Now, solve the equation as shown,

  $\begin{array}{c}\frac{\mathrm{k}}{\mathrm{k}+1}+\frac{\mathrm{k}}{\mathrm{k}-2}=2\\ \frac{{\mathrm{k}}^2-2\mathrm{k}+{\mathrm{k}}^2+\mathrm{k}}{\left(\mathrm{k}+1\right)\left(\mathrm{k}-2\right)}=2\\ 2{\mathrm{k}}^2-\mathrm{k}=2{\mathrm{k}}^2-2\mathrm{k}-4\\ 2{\mathrm{k}}^2-\mathrm{k}-2{\mathrm{k}}^2+2\mathrm{k}=-4\\ \mathrm{k}=-4\end{array}$

Now, check the result by back substitution.

  $\begin{array}{c}\frac{\mathrm{k}}{\mathrm{k}+1}+\frac{\mathrm{k}}{\mathrm{k}-2}=2\\ \frac{-4}{-4+1}+\frac{-4}{-4-2}=2\\ \frac{4}{3}+\frac{4}{6}=2\\ \frac{12}{6}=2\\ 2=2\end{array}$

Which is true.

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