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Simplify the following rational expression: $5/(x+3) - 7x/(x-1)$

I came across this question in my homework and because it is a fraction, I decided that I needed to establish a common denominator of: $$x-3$$ But how do I convert x - 1 into x-3 ?

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  • $\begingroup$ Why a common denominator of $x-3$? $\endgroup$ Oct 1, 2015 at 0:38
  • $\begingroup$ I don't see how that expression can be simplified. It can be expressed as a single term with denominator $(x+3)(x-1)$, but I wouldn't call that simpler. $\endgroup$
    – TonyK
    Oct 1, 2015 at 0:41

1 Answer 1

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Notice, we can simplify the given expression as follows $$\frac{5}{x+3}-\frac{7x}{x-1}$$ $$=\frac{5}{x+3}-\frac{7x-7+7}{x-1}$$ $$=\frac{5}{x+3}-\frac{7(x-1)+7}{x-1}$$ $$=\frac{5}{x+3}-\frac{7(x-1)}{x-1}-\frac{7}{x-1}$$ $$=\frac{5}{x+3}-7-\frac{7}{x-1}$$ $$=-7+\frac{5}{x+3}-\frac{7}{x-1}$$

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Using partial fraction $$\frac{5}{x+3}-\frac{7x}{x-1}$$ $$=\frac{5(x-1)-7x(x+3)}{(x+3)(x-1)}$$ $$=\frac{5x-5-7x^2-21x}{(x+3)(x-1)}$$ $$=\frac{-7x^2-16x-5}{(x+3)(x-1)}$$ $$=\frac{-(7x^2+16x+5)}{(x+3)(x-1)}$$ $$=-7-\frac{2x+26}{(x+3)(x-1)}$$ $$=-7-\frac{2(x+13)}{(x-1)(x-3)}$$ Making partial fractions of $\frac{x+13}{(x-1)(x-3)}=\frac{A}{x-1}+\frac{B}{x+3}\implies A=\frac{7}{2}, \ B=-\frac{5}{2}$, we get $$\color{blue}{\frac{5}{x+3}-\frac{7x}{x-1}}=\color{red}{-7-\frac{7}{x-1}+\frac{5}{x+3}}$$

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