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I know most of the people on this forum/site are very advanced, and I'm just sitting here wondering how in the world you do this equation, or "simplify" it.

I know how to do equations like these when there are only two terms, as in (a/b + c/d), but I can't figure out why the same rule doesn't apply to (a/b + c/d + e/f), can someone please help me, I've searched everywhere, I can't wait a whole day to ask my teacher, I need to do this, every where I search they just give me 2 term equations, thanks in advanced, sorry for being so un-professional.

https://i.stack.imgur.com/OlYfx.jpg

Sam Chahine
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1 Answers1

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You can think of the addition of adding $3$ fractions $\frac{a}{b}+\frac{c}{d}+\frac{e}{f}$, thanks to the associative property of addition, as $\left(\frac{a}{b}+\frac{c}{d}\right)+\frac{e}{f}$, so you just need to apply the rule for adding two fractions twice: $$ \left(\frac{a}{b}+\frac{c}{d}\right)+\frac{e}{f}=\frac{ad+bc}{bd}+\frac{e}{f}=\frac{(ad+bc)f+bde}{bdf}=\frac{adf+bcf+bde}{bdf} $$

As an example, Your question gives $$\begin{align}&\frac{1}{x^2-7x+10}-\frac{2}{x^2-2x-15}+\frac{4}{x^2+x-6}\\ =&\frac{1}{(x-5)(x-2)}-\frac{2}{(x-5)(x+3)}+\frac{4}{(x+3)(x-2)} \\ =&\frac{1}{x-5}\left(\frac{1}{x-2}-\frac{2}{x+3}\right)+\frac{4}{(x+3)(x-2)} \\ =&\frac{1}{x-5}\left(\frac{x+3-2(x-2)}{(x-2)(x+3)}\right)+\frac{4}{(x+3)(x-2)} \\ =&\frac{7-x}{(x-5)(x-2)(x+3)}+\frac{4}{(x+3)(x-2)} \\ =&\frac{1}{(x-2)(x+3)}\left(\frac{7-x}{x-5}+4\right) \\ =&\frac{1}{(x-2)(x+3)}\left(\frac{7-x+4(x-5)}{x-5}\right) \\ =&\frac{3x-13}{(x-5)(x-2)(x+3)} \end{align}$$

Tim Ratigan
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