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}\xa03(2×2\xa0- 14x + 22) = 10(x2\xa0- 8x + 15) [ by cross multiplication method ]{tex}\\Rightarrow{/tex}\xa06x2 – 42x + 66 = 10×2 – 80x +150{tex}\\Rightarrow{/tex}\xa04x2 – 38x + 84 = 0 {tex}\\Rightarrow{/tex}\xa02x2\xa0- 19x + 42 = 0{tex}\\Rightarrow{/tex}\xa02x2\xa0- 12x – 7x + 42 = 0 {tex}\\Rightarrow{/tex}\xa02x(x -6 ) – 7(x\xa0- 6) = 0{tex}\\Rightarrow{/tex}\xa0(x – 6)(2x – 7) = 0 {tex}\\Rightarrow{/tex}\xa0x – 6 = 0 or 2x – 7 = 0 Either x = 6 or x =\xa0{tex}\\frac{7}{2}{/tex}.Hence, the roots of given equation are\xa06 and {tex}\\frac{7}{2}{/tex}.