Question

he tiles to the correct boxes to complete the pairs.
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

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

Step2: Expand numerator

$$\frac{x^2 -x +5x -5 -2}{x-1} = \frac{x^2 +4x -7}{x-1}$$

Step3: Combine into single fraction

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

Step4: Expand numerator

$$\frac{x^2 -2x +1 +6}{x-1} = \frac{x^2 -2x +7}{x-1}$$

Step5: Combine into single fraction

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

Step6: Expand numerator

$$\frac{2x^2 -2x +x -1 -6}{x-1} = \frac{2x^2 -x -7}{x-1}$$

Step7: Combine into single fraction

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

Step8: Expand numerator

$$\frac{2x^2 -2x -x +1 +6}{x-1} = \frac{2x^2 -3x +7}{x-1}$$

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}$