Question

A classmate simplified a rational expression below

Answer 1

Part a)

Explain the error in this simplification.

Given the simplified expression

$1-\frac{2}{x-2}=\frac{x+1}{x+2}$

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

$\:1-2x-4=x^2-x-2$

$-2x-3=x^2-x-2\:\:$

$0=x^2-2x-2\:\:$

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

$x=1$

Identifying the Main Error

$1-\frac{2}{x-2}=\frac{x+1}{x+2}$

$1-2\left(x+2\right)=\left(x+1\right)\left(x-2\right)$   ← ERROR Starts here

Here is the Explanation of the Error

$\mathrm{The\:equation\:should\:have\:been\:Multiplied\:by\:LCM=}\left(x-2\right)\left(x+2\right)$. In your case you wrongly multiply the equation.

CORRECTION

HERE IS HOW YOU SHOULD HAVE MULTIPLIED BY LCM = (x-2)(x+2):

$1-\frac{2}{x-2}=\frac{x+1}{x+2}$

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

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

Part b)

Show your work as you correct the error

Here is the complete correction of the error.

Considering the expression

$1-\frac{2}{x-2}=\frac{x+1}{x+2}$

$\mathrm{Find\:Least\:Common\:Multiplier\:of\:}x-2,\:x+2:\quad \left(x-2\right)\left(x+2\right)$

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

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

$x^2-2x-8=x^2-x-2$

$x^2-2x-8+8=x^2-x-2+8$

$x^2-2x=x^2-x+6$

$-x=6$

$\frac{-x}{-1}=\frac{6}{-1}$

$x=-6$