Solve 1 by x-3 + 2 …

We have,
{tex}\frac{1}{{x - 3}} + \frac{2}{{x - 2}} = \frac{8}{x}{/tex}
{tex}\Rightarrow \frac{{1(x - 2) + 2(x - 3)}}{{(x - 3)(x - 2)}} = \frac{8}{x}{/tex}
{tex}\Rightarrow \frac{{x - 2 + 2x - 6}}{{(x - 3)(x - 2)}} = \frac{8}{x}{/tex}
{tex}\Rightarrow \frac{{3x - 8}}{{{x^2} - 2x - 3x + 6}} = \frac{8}{x}{/tex}
{tex}\Rightarrow \frac{{3x - 8}}{{{x^2} - 5x + 6}} = \frac{8}{x}{/tex}

Cross multiply,
{tex}\Rightarrow{/tex} {tex}x(3x-8)=8(x^2-5x+6){/tex}
{tex}\Rightarrow{/tex} {tex}3x^2-8x=8x^2-40x+48{/tex}
{tex}\Rightarrow{/tex} {tex}8x^2-40x+48-3x^2+8x=0{/tex}
{tex}\Rightarrow{/tex} {tex}5x^2-32x+48=0{/tex}

Factorise the equation,
{tex}\Rightarrow{/tex} 5x² - 20x - 12x + 48 = 0
{tex}\Rightarrow{/tex} 5x(x - 4) - 12(x - 4) = 0
{tex}\Rightarrow{/tex} (5x - 12) (x - 4) = 0
{tex}\Rightarrow{/tex} (5x - 12) = 0 or (x - 4) = 0
{tex} \Rightarrow x = \frac{{12}}{5}{/tex} or x = 4