Chapter 5.6, Problem 10Q

Solution

To calculate: Solve $\frac{3x+6}{x^2-4}=\frac{x+1}{x-2}$ equation and check extraneous solution. $$

Solution of $\frac{3x+6}{x^2-4}=\frac{x+1}{x-2}$ is $x=\pm 2$.

Given information: Given $\frac{3x+6}{x^2-4}=\frac{x+1}{x-2}$ . Where x is variable.

Calculation: Solution of:

  $\frac{3x+6}{x^2-4}=\frac{x+1}{x-2}$

  $\frac{3x+6}{\left(x-2\right)\left(x+2\right)}=\frac{x+1}{x-2}$ (Using Formula $a^2-b^2=\left(a-b\right)\left(a+b\right)$ )

Cut $\left(x-2\right)$ on both side in above equation.

  $\frac{3x+6}{\left(x+2\right)}=\frac{x+1}{1}$

By using cross multiplication solve this equation

Now,

  $\left(3x+6\right)=\left(x+1\right)\times \left(x+2\right)$

  $\left(3x+6\right)=\left(x^2+2x+x+2\right)$

  $\left(3x+6\right)=\left(x^2+3x+2\right)$

Subtract 3 x on both side in above equation

  $\left(3x+6-3x\right)=\left(x^2+3x+2-3x\right)$

  $6=\left(x^2+2\right)$

  $x^2=4$

  $x=\pm 2$

Thus Solution of $\frac{3x+6}{x^2-4}=\frac{x+1}{x-2}$

  $x=\pm 2$.

Now check this solution is extraneous or not.

Put this value in original equation $\frac{3x+6}{x^2-4}=\frac{x+1}{x-2}$

  $\frac{3\left(2\right)+6}{{\left(2\right)}^2-4}=\frac{2+1}{2-2}$

  $\frac{3\left(-2\right)+6}{{\left(-2\right)}^2-4}=\frac{-2+1}{-2-2}$

  $\frac{6+6}{4-4}=\frac{2+1}{2-2}$

  $\frac{+6-6}{4-4}=\frac{-2+1}{-2-2}$

  $\frac{6+6}{0}=\frac{2+1}{0}$

  $\frac{-6+6}{0}=\frac{-2+1}{0}$

  $0=0$

  $0=0$

Hence $x=\pm 2$ is not extraneous solution.