101
Solve for $x$: $\frac{x-2}{x-3} + \frac{x-4}{x-5} = \frac{10}{3}$; $x \neq 3,5$
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Solution:
(a) $\frac{x-2}{x-3} + \frac{x-4}{x-5} = \frac{10}{3}$
$\Rightarrow 3[(x - 2) (x – 5) + (x – 3) (x − 4)] = 10 (x – 3) (x - 5)$ (1.5 Mark)
$\Rightarrow 4x^2 - 38x + 84 = 0$ or $2x^2 - 19x + 42 = 0$ (1.5 Mark)
$\Rightarrow (x – 6) (2x – 7) = 0$ (1 Mark)
$\Rightarrow x = 6, x = \frac{7}{2}$ (1 Mark)
(a) $\frac{x-2}{x-3} + \frac{x-4}{x-5} = \frac{10}{3}$
$\Rightarrow 3[(x - 2) (x – 5) + (x – 3) (x − 4)] = 10 (x – 3) (x - 5)$ (1.5 Mark)
$\Rightarrow 4x^2 - 38x + 84 = 0$ or $2x^2 - 19x + 42 = 0$ (1.5 Mark)
$\Rightarrow (x – 6) (2x – 7) = 0$ (1 Mark)
$\Rightarrow x = 6, x = \frac{7}{2}$ (1 Mark)