Chapter 11.5, Problem 16P

Solve each equation. Check your solutions. If there is no solution, write no solution.

$\frac{t-1}{3t^2-t-2}-\frac{2t-3}{3t+2}=\frac{-4}{2t-2}$

To determine

To solve the equation and check the solution.

Answer to Problem 16P

The solution is $t=0,6$ .

Explanation of Solution

Given information:

The given equation is:

  $\frac{t-1}{3t^2-t-2}-\frac{2t-3}{3t+2}=\frac{-4}{2t-2}$

Calculation:

The solution can be obtained by:

  $\begin{array}{l}\frac{t-1}{3t^2-t-2}-\frac{2t-3}{3t+2}=\frac{-4}{2t-2}\\ \frac{t-1}{3t^2-3t+2t-2}-\frac{2t-3}{3t+2}=\frac{-4}{2t-2}\\ \frac{t-1}{3t\left(t-1\right)+2\left(t-1\right)}-\frac{2t-3}{3t+2}=\frac{-4}{2t-2}\\ \frac{t-1}{\left(t-1\right)\left(3t+2\right)}-\frac{2t-3}{3t+2}=\frac{-4}{2t-2}\\ \frac{1}{\left(3t+2\right)}-\frac{2t-3}{3t+2}=\frac{-4}{2\left(t-1\right)}\\ \frac{1-2t+3}{\left(3t+2\right)}=\frac{-2}{t-1}\\ \frac{4-2t}{\left(3t+2\right)}=\frac{-2}{t-1}\\ \frac{2\left(2-t\right)}{\left(3t+2\right)}=\frac{-2}{t-1}\\ \frac{\left(2-t\right)}{\left(3t+2\right)}=\frac{-2}{2\left(t-1\right)}\\ \frac{\left(2-t\right)}{\left(3t+2\right)}=\frac{-1}{t-1}\\ \left(2-t\right)\left(t-1\right)=-1\left(3t+2\right)\\ 2t-2-t^2+t=-3t-2\\ 3t-2-t^2+3t+2=0\\ -t^2+6t=0\\ t^2-6t=0\\ t\left(t-6\right)=0\\ t=0,6\end{array}$

Hence, the solution is $t=0,6$ .

Put the value of $t$ in the solution to check the solution.