How do you solve $\frac{x - 2}{x - 3} + \frac{x - 3}{x - 2} = \frac{2 x^{2}}{x^{2} - 5 x + 6}$ $(x-2)/(x-3)+(x-3)/(x-2)=(2x^2)/(x^2-5x+6)$ ?

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How do you solve $\frac{x - 2}{x - 3} + \frac{x - 3}{x - 2} = \frac{2 x^{2}}{x^{2} - 5 x + 6}$ $(x-2)/(x-3)+(x-3)/(x-2)=(2x^2)/(x^2-5x+6)$ ?

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Answer 1 · 2016-11-23T14:44:24

$x = \frac{13}{10}$ $x = 13/10$

Explanation:

First, you must get each fraction over a common denominator which is $x^{2} - 5 x + 6$ $x^2 - 5x + 6$ :

$\frac{x - 2}{x - 2} \frac{x - 2}{x - 3} + \frac{x - 3}{x - 3} \frac{x - 3}{x - 2} = \frac{2 x^{2}}{x^{2} - 5 x + 6}$ $(x - 2)/(x - 2) (x - 2)/(x - 3) + (x - 3)/(x - 3) (x - 3)/(x - 2) = (2x^2)/(x^2 - 5x + 6)$

$\frac{x^{2} - 4 x + 4}{x^{2} - 5 x + 6} + \frac{x^{2} - 6 x + 9}{x^{2} - 5 x + 6} = \frac{2 x^{2}}{x^{2} - 5 x + 6}$ $(x^2 - 4x + 4)/(x^2 - 5x + 6) + (x^2 - 6x + 9)/(x^2 - 5x + 6) = (2x^2)/(x^2 - 5x + 6)$

Next, add the fractions on the left side of the equation:

$\frac{(x^{2} - 4 x + 4) + (x^{2} - 6 x + 9)}{x^{2} - 5 x + 6} = \frac{2 x^{2}}{x^{2} - 5 x + 6}$ $((x^2 - 4x + 4) + (x^2 - 6x + 9))/(x^2 - 5x + 6) = (2x^2)/(x^2 - 5x + 6)$

$\frac{2 x^{2} - 10 x + 13}{x^{2} - 5 x + 6} = \frac{2 x^{2}}{x^{2} - 5 x + 6}$ $(2x^2 - 10x + 13)/(x^2 - 5x + 6) = (2x^2)/(x^2 - 5x + 6)$

We can now multiply each side of the equation by $x^{2} - 5 x + 6$ $x^2 - 5x + 6$ to eliminate the fraction:

$(x^{2} - 5 x + 6) \frac{2 x^{2} - 10 x + 13}{x^{2} - 5 x + 6} = (x^{2} - 5 x + 6) \frac{2 x^{2}}{x^{2} - 5 x + 6}$ $(x^2 - 5x + 6) (2x^2 - 10x + 13)/(x^2 - 5x + 6) = (x^2 - 5x + 6) (2x^2)/(x^2 - 5x + 6)$

$cancel((x^2 - 5x + 6)) (2x^2 - 10x + 13)/cancel((x^2 - 5x + 6)) = cancel((x^2 - 5x + 6)) (2x^2)/cancel((x^2 - 5x + 6))$

$2x^2 - 10x + 13 = 2x^2$

We can now solve for $x$ :

$2x^2 - 10x + 13 - 2x^2= 2x^2 - 2x^2$

$-10x + 13 = 0$

$-10x + 13 - 13 = 0 - 13$

$-10x = -13$

$(-10x)/-10 = (-13)/(-10)$

$x = 13/10$

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How do you solve $\frac{x - 2}{x - 3} + \frac{x - 3}{x - 2} = \frac{2 x^{2}}{x^{2} - 5 x + 6}$ $(x-2)/(x-3)+(x-3)/(x-2)=(2x^2)/(x^2-5x+6)$ ?

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Explanation:

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How do you solve $\frac{x - 2}{x - 3} + \frac{x - 3}{x - 2} = \frac{2 x^{2}}{x^{2} - 5 x + 6}$ $(x-2)/(x-3)+(x-3)/(x-2)=(2x^2)/(x^2-5x+6)$ ?