Chapter 8, Problem 29CLR

To determine

To find: The solution for the given equation.

Answer to Problem 29CLR

The solution for the equation is $\left(0,\frac{1}{3}\right)$ .

Explanation of Solution

Given information:

The given equation is $\frac{2z}{z-3}-\frac{1}{z+1}=-1$ .

Calculation:

Calculate the solution for equation.

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

  $\begin{array}{c}3z^2-z=0\\ z\left(3z-1\right)=0\end{array}$

The solutions are.

  $\begin{array}{c}z=0\\ \left(3z-1\right)=0\\ 3z=1\\ z=\frac{1}{3}\end{array}$

The solution set is $\left(0,\frac{1}{3}\right)$ .

Therefore, the solution for the equation is $\left(0,\frac{1}{3}\right)$ .