How do you solve (x-2)/(x-3)+(x-3)/(x-2)=(2x^2)/(x^2-5x+6)?

Nov 23, 2016

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

Explanation:

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

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

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

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

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

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

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

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

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

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

We can now solve for $x$:

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

$- 10 x + 13 = 0$

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

$- 10 x = - 13$

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

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