How do you solve $\frac{2 x - 3}{x - 1} - \frac{x - 1}{x + 2} = \frac{2 x - 5}{x + 2} + \frac{2 - x}{1 - x}$ $(2x-3) /( x-1) - (x-1 )/( x+2) = (2x-5) /( x+2) + (2-x) /( 1-x)$ ?

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How do you solve $\frac{2 x - 3}{x - 1} - \frac{x - 1}{x + 2} = \frac{2 x - 5}{x + 2} + \frac{2 - x}{1 - x}$ $(2x-3) /( x-1) - (x-1 )/( x+2) = (2x-5) /( x+2) + (2-x) /( 1-x)$ ?

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Answer 1 · 2015-10-18T00:12:21

$x = \frac{8}{5}$ $x = 8/5$

Explanation:

Mutiply both sides by $(x - 1) (x + 2)$ $(x-1)(x+2)$

$(2 x - 3) (x + 2) - {(x - 1)}^{2} = (2 x - 5) (x - 1) + (x - 2) (x + 2)$ $(2x-3)(x+2) - (x-1)^2 = (2x - 5)(x-1) +(x-2)(x+2)$

Simplify

$x^{2} + 3 x - 7 = 3 x^{2} - 7 x + 1$ $x^2 + 3x - 7 = 3x^2 - 7x + 1$
$2 x^{2} - 10 x + 8 = 0$ $2x^2 - 10x + 8 = 0$
$x^{2} - 5 x + 4 = 0$ $x^2 - 5x + 4 = 0$
$(x - 1) (x - 4) = 0$ $(x - 1)(x - 4) = 0$
$x = 1 or x = 4$ $x = 1 or x = 4$

$x$ $x$ must differ from $1$ $1$ or we'd have a division by zero, so $x = 0$ $x = 0$

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How do you solve $\frac{2 x - 3}{x - 1} - \frac{x - 1}{x + 2} = \frac{2 x - 5}{x + 2} + \frac{2 - x}{1 - x}$ $(2x-3) /( x-1) - (x-1 )/( x+2) = (2x-5) /( x+2) + (2-x) /( 1-x)$ ?

1 Answer

Explanation:

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How do you solve 2x−3x−1−x−1x+2=2x−5x+2+2−x1−x(2x-3) /( x-1) - (x-1 )/( x+2) = (2x-5) /( x+2) + (2-x) /( 1-x) ?

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How do you solve $\frac{2 x - 3}{x - 1} - \frac{x - 1}{x + 2} = \frac{2 x - 5}{x + 2} + \frac{2 - x}{1 - x}$ $(2x-3) /( x-1) - (x-1 )/( x+2) = (2x-5) /( x+2) + (2-x) /( 1-x)$ ?