Chapter 5.6, Problem 12Q

Solution

  

To calculate: Solve $\frac{x-3}{x+2}=\frac{x-1}{3x-1}$ equation and check extraneous solution.

Solution of $\frac{x-3}{x+2}=\frac{x-1}{3x-1}$ is $x=5$ and $x=\frac{1}{2}$.

Given information: Given $\frac{x-3}{x+2}=\frac{x-1}{3x-1}$ . Where x is variable.

Calculation: Solution of:

  $\frac{x-3}{x+2}=\frac{x-1}{3x-1}$

By using cross multiplication solve this equation

Now,

  $\left(x-3\right)\left(3x-1\right)=\left(x-1\right)\left(x+2\right)$

  $\left(3x^2-x-9x+3\right)=\left(x^2+2x-x-2\right)$

  $\left(3x^2-10x+3\right)=\left(x^2+x-2\right)$

  $\left(3x^2-10x+3\right)-\left(x^2+x-2\right)=0$

  $\left(\left(3x^2-x^2\right)+\left(-10x-x\right)+\left(3+2\right)\right)=0$

  $2x^2-11x+5=0$

Now solve this quadratic equation. By splitting middle term

  $2x^2-10x-x+5=0$

  $\left(2x^2-10x\right)+\left(-x+5\right)=0$

  $2\left(x-5\right)-1\left(x-5\right)=0$

  $\left(x-5\right)\left(2x-1\right)=0$

  $x=5$ And $x=\frac{1}{2}$

Thus Solution of $\frac{x-3}{x+2}=\frac{x-1}{3x-1}$ is $x=5$ and $x=\frac{1}{2}$

Now check this solution is extraneous or not.

Put this value $x=5$ and $x=\frac{1}{2}$ in original equation $\frac{x-3}{x+2}=\frac{x-1}{3x-1}$

  $\frac{5-3}{5+2}=\frac{5-1}{3\left(5\right)-1}$

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

  $\frac{2}{7}=\frac{2}{7}$

  $-1=-1$

Hence $x=5$ and $x=\frac{1}{2}$ is not extraneous solution.