How do you solve $\frac{14}{x - 8} + 2 = \frac{10}{x - 8}$ $14/(x - 8) + 2 = 10/(x - 8)$ ?

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Answer 1 · 2015-11-21T11:04:16

$x = 6$ $x=6$

Assuming $x \neq 8$ $x ne 8$ , we can convert the expression into

$\frac{14}{x - 8} + \frac{2 (x - 8)}{x - 8} = \frac{10}{x - 8}$ $\frac{14}{x-8} + \frac{2(x-8)}{x-8} = \frac{10}{x-8}$

And thus

$\frac{14+2(x-8)}{cancel(x-8)} = \frac{10}{cancel(x-8)}$

Expanding the first numerator, we have $14+2x-16 = 2x-2$

So, the equation becomes $2x-2 = 10$ , and finally
$2x=12 -> x=6$

How do you solve 14/(x - 8) + 2 = 10/(x - 8)14x−8+2=10x−814/(x - 8) + 2 = 10/(x - 8)?