Question

x-1 upon x-2+x-3 upon x-4 =10 upon 3

Answer 1

$\frac{x - 1}{x - 2} + \frac{x - 3}{ x - 4} = \frac{10}{3} \\ \\ solving \: by \: taking \: lcm \\ \\ \frac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \frac{10}{3} \\ \\ \frac{ {x}^{2} - 4x - x + 4 + {x}^{2} - 2x - 3x + 6}{ {x }^{2} - 4x - 2x + 8} = \frac{10}{3} \\ \\ \frac{ {2x}^{2} - 10x + 10}{ {x}^{2} - 6x + 8} = \frac{10}{3} \\ \\ \frac{ {x}^{2} - 5x + 5}{ {x}^{2} - 6x + 8} = \frac{5}{3} \\ \\ 3 {x}^{2} - 15x + 15 = 5 {x}^{2} - 30x + 40 \\ \\ 2 {x}^{2} - 15x + 25 = 0 \\ \\ 2 {x}^{2} - 10x - 5x + 25 = 0 \\ \\ 2x(x - 5) - 5(x - 5) = 0 \\ \\ (x - 5) \: (2x - 5) = 0 \\ \\ x = 5 \: \: \: \: \: or \: \: \: x = \frac{5}{2}$

Hope it helps.

Answer 2

ANSWER....

IN SUCH CASE WHEN VARIABLE IS IN ONE SIDE AND CONSTANT IS IN ANOTHER SIDE WE CAN SHIFT THE VARIABLE SO THAT THE CALCULATION BECOME EASIER....

$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 10 \\ \\ shift \: \: the \: \: variable \\ \\ \frac{x - 1}{x - 2} = 10 - \frac{x - 3}{x - 4} \\ \\ \frac{x - 1}{x - 2} = \frac{10(x - 4) - (x - 3)}{x - 4} \\ \\ \frac{x - 1}{x - 2} = \frac{10x - 40 - x + 3}{x - 4} \\ \\ \frac{x - 1}{x - 2} = \frac{9x - 37}{x - 4} \\ \\ (x - 1)(x - 4) = (x - 2)(9x - 37) \\ \\ {x}^{2} - 5x + 4 = 9 {x}^{2} - 18x - 74x + 74 \\ \\ 8 {x}^{2} - 87x + 70 = 0 \\ \\ 8 {x}^{2} - 80x - 7x + 70 = 0 \\ \\ 8x(x - 10) - 7(x - 10) = 0 \\ \\ (8x - 7)(x - 10) = 0 \\ \\ from \: here \: we \: get \\ \\ x = \frac{7}{8} \: \: or \: \: 10$

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