Question

$\frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = \frac{10}{3}$
solve this please!! I need help!!!
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Answer 1

Answer:

$x = 8$

Step-by-step explanation:

$\frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = \frac{10}{3} \\ \frac{(x - 4)(x -7 ) + (x - 6)(x - 5)}{(x - 5)(x - 7)} = \frac{10}{3} \\ \frac{x(x - 7) - 4(x - 7) + x(x - 5) - 6(x - 5)}{x(x - 7) - 5(x - 7)} = \frac{10}{3} \\ \frac{ {x}^{2} - 7x - 4x + 28 + {x}^{2} - 5x - 6x + 30 }{ {x}^{2} - 7x - 5x + 35 } = \frac{10}{3} \\ \frac{ {x}^{2} + {x}^{2} - 7x - 4x - 5x - 6x + 28 + 30 }{ {x}^{2} - 7x - 5x + 35 } = \frac{10}{3} \\ \frac{2 {x}^{2} - 22x + 58 }{ {x}^{2} - 12x + 35} = \frac{10}{3} \\ Cross \: multiply \\ 10( {x}^{2} - 12x + 35) = 3(2 {x}^{2} - 22x + 58) \\ 10 {x}^{2} - 120x + 350 = 6 {x}^{2} - 66x + 174 \\ 10 {x}^{2} - 6 {x}^{2} - 120x + 66x + 350 - 174 = 0 \\ 4 {x}^{2} - 54x + 176 = 0 \\ 2 {x}^{2} - 27x + 88 = 0 \\ 2 {x}^{2} - 16x - 11x + 88 = 0 \\ (2 {x}^{2} - 16x) - (11x + 88) = 0 \\ 2x(x - 8) - 11(x - 8) = 0 \\ (x - 8)(2x - 11) = 0 \\ (x - 8 = 0)(2x - 11 = 0) \\ x = 8 \: and \: x = \frac{11}{2} \\ Therefore, \: x = 8$