Question

drag the tiles to the correct boxes to complete the pairs.
match the rational expressions to their rewritten forms.
$(x + 5) + \frac{-2}{x - 1}$
$(x - 1) + \frac{6}{x - 1}$
$(2x + 1) + \frac{-6}{x - 1}$
$(2x - 1) + \frac{6}{x - 1}$

Explanation:

Step1: Combine into single fraction

For $(x + 5) + \frac{-2}{x-1}$, rewrite $x+5$ with denominator $x-1$:
$\frac{(x+5)(x-1)}{x-1} + \frac{-2}{x-1} = \frac{x^2 +4x -5 -2}{x-1} = \frac{x^2 +4x -7}{x-1}$

Step2: Combine into single fraction

For $(x - 1) + \frac{6}{x-1}$, rewrite $x-1$ with denominator $x-1$:
$\frac{(x-1)(x-1)}{x-1} + \frac{6}{x-1} = \frac{x^2 -2x +1 +6}{x-1} = \frac{x^2 -2x +7}{x-1}$

Step3: Combine into single fraction

For $(2x + 1) + \frac{-6}{x-1}$, rewrite $2x+1$ with denominator $x-1$:
$\frac{(2x+1)(x-1)}{x-1} + \frac{-6}{x-1} = \frac{2x^2 -2x +x -1 -6}{x-1} = \frac{2x^2 -x -7}{x-1}$

Step4: Combine into single fraction

For $(2x - 1) + \frac{6}{x-1}$, rewrite $2x-1$ with denominator $x-1$:
$\frac{(2x-1)(x-1)}{x-1} + \frac{6}{x-1} = \frac{2x^2 -2x -x +1 +6}{x-1} = \frac{2x^2 -3x +7}{x-1}$

Answer:

1. $(x + 5) + \frac{-2}{x-1}

ightarrow \frac{x^2 +4x -7}{x-1}$

1. $(x - 1) + \frac{6}{x-1}

ightarrow \frac{x^2 -2x +7}{x-1}$

1. $(2x + 1) + \frac{-6}{x-1}

ightarrow \frac{2x^2 -x -7}{x-1}$

1. $(2x - 1) + \frac{6}{x-1}

ightarrow \frac{2x^2 -3x +7}{x-1}$