X-2/x-3+x-4/x-5=10/3

The given equation is:
{tex}\frac { x - 2 } { x - 3 } + \frac { x - 4 } { x - 5 } = \frac { 10 } { 3 }{/tex}

By cross multiplication method we have,
{tex} \frac { ( x - 2 ) ( x - 5 ) + ( x - 4 ) ( x - 3 ) } { ( x - 3 ) ( x - 5 ) } = \frac { 10 } { 3 }{/tex}
{tex}\Rightarrow \frac { \left( x ^ { 2 } - 7 x + 10 \right) + \left( x ^ { 2 } - 7 x + 12 \right) } { \left( x ^ { 2 } - 8 x + 15 \right) } = \frac { 10 } { 3 }{/tex}
{tex}\Rightarrow \frac { \left( 2 x ^ { 2 } - 14 x + 22 \right) } { \left( x ^ { 2 } - 8 x + 15 \right) } = \frac { 10 } { 3 }{/tex}
{tex}\Rightarrow{/tex} 3(2x² - 14x + 22) = 10(x² - 8x + 15) [ by cross multiplication method ]
{tex}\Rightarrow{/tex} 6x² - 42x + 66 = 10x² - 80x +150
{tex}\Rightarrow{/tex} 4x² - 38x + 84 = 0 {tex}\Rightarrow{/tex} 2x² - 19x + 42 = 0
{tex}\Rightarrow{/tex} 2x² - 12x - 7x + 42 = 0 {tex}\Rightarrow{/tex} 2x(x -6 ) - 7(x - 6) = 0
{tex}\Rightarrow{/tex} (x - 6)(2x - 7) = 0 {tex}\Rightarrow{/tex} x - 6 = 0 or 2x - 7 = 0
   Either x = 6 or x = {tex}\frac{7}{2}{/tex}.
Hence, the roots of given equation are 6 and {tex}\frac{7}{2}{/tex}.