Chapter 5.9, Problem 25WE

To determine

To solve and check , the given equation has no solution of $\frac{t}{t-1}=\frac{1}{t+2}+\frac{3}{t^2+t-2}$

Answer to Problem 25WE

No solution

Explanation of Solution

Given that: : $\frac{t}{t-1}=\frac{1}{t+2}+\frac{3}{t^2+t-2}$

First factor the $t^2+t-2$

  $\begin{array}{c}{t}^2+t-2=t^2+(2-1)t-2\\ =t^2+2t-t-2\\ =t(t+2)-1(t+2)\\ =(t-1)(t+2)\end{array}$

Rewrite the given equation

  $\frac{t}{t-1}=\frac{1}{t+2}+\frac{3}{(t-1)(t+2)}$

Multiply by $(t+2)(t-1)$ , LCM of $(t+2)and(t-1)$

  $(t-1)(t+2)\left(\frac{t}{t-1}\right)=(t+2)(t-1)\left(\frac{1}{t+2}+\frac{3}{(t-1)(t+2)}\right)$

Apply distributive property

  $\begin{array}{l}t(t+2)=(t-1)+3\\ t^2+2t=t+2\\ t^2+t-2=0\end{array}$

Apply the factor method and Apply zero product property to find the solution of equation

  $\begin{array}{c}{t}^2+t-2=t^2+(2-1)t-2\\ =t^2+2t-t-2\\ =t(t+2)-1(t+2)\\ =(t-1)(t+2)\\ t=1,-2\end{array}$

When    t  equal to -2 and 1, a denominator in the original equation has a value of 0. Therefore,

-2 and 1 are not in the domain of y and cannot be solution.

Factor value exactly same of denominator therefore, the solution is undefined.

Check:

  $\begin{array}{l}Fort=-2\\ \frac{-2}{-2-1}\ne \frac{1}{-2+2}+\frac{3}{4-2-2}\to undefined\\ Similarly\\ Fort=1\\ \frac{1}{1-1}\ne \frac{1}{1+2}+\frac{3}{1+1-2}\to undefined\end{array}$

Hence, No solution of equation.