Finding LCM of an expression

Question

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I am currently trying to figure out how to solve this problem. I know how they got the denominator but am unsure how they got 6x in the numerator? Can somebody explain this to me?

Answer 1

$$\frac{x}{x^2 - 8x + 7} - \frac{x}{x^2-2x-35} =$$

$$\frac{x}{(x-7)(x-1)} - \frac{x}{(x-7)(x+5)}=$$

$$\frac{x(x+5) - x(x-1)}{(x-7)(x-1)(x+5)}=$$

$$\frac{x^2+5x -x^2 +x}{(x-7)(x-1)(x+5)}=$$

$$\frac{6x}{(x-7)(x-1)(x+5)}$$

Answer 2

Factor both denominators; $x^2 - 8x + 7$ factors as $(x - 1)(x - 7)$, and $x^2 - 2x - 35$ factors as $(x - 7)(x + 5)$. So $(x - 1)(x - 7)(x + 5)$ is a multiple of both. Now, to get each fraction to have that denominator, you need to multiply by the extra bits; for example, for $\frac{x}{x^2 - 8x + 7}$, you need to multiply the top and bottom by $x + 5$, because that's the missing part in the bottom. That gives you $\frac{x^2 + 5x}{(x - 7)(x - 1)(x + 5)}$. For the other one, the missing piece is $x - 1$, so you get $\frac{x^2 - x}{(x - 7)(x - 1)(x + 5)}$. Then subtracting the numerators gives you $x^2 + 5x - (x^2 - x) = 6x$.