How could I simplify $\dfrac 8{x+5} - \dfrac 3{5-x}$? I know we should find the LCM for the denominators and then simplify, but this question is little confusing.
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1What do you think that the LCM for the denominators is? – paw88789 Jul 09 '19 at 14:34
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Are you familiar with the method of multiplying by $1$? In this case, you might multiply $\dfrac 8{x+5}$ by $\dfrac{5-x}{5-x}$ (which is also equal to $\dfrac {x-5}{x-5}$), and similar for the other term, then combine... – abiessu Jul 09 '19 at 14:34
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Why you think that the answer is little confusing? – ESCM Jul 09 '19 at 14:35
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@SFuard You have been asking many elementary questions here, sometimes the same one several times. Please work harder on your own before coming for help. When you do get an answer to a question you should accept it (the check mark) and upvote it (the up arrow). You can upvote more than one answer to your question or to any answer to any other question. – Ethan Bolker Jul 11 '19 at 13:06
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It is $$\frac{8}{x+5}-\frac{3}{5-x}=\frac{8(5-x)-3(5+x)}{(5+x)(5-x)}$$ for $$x\neq \pm5$$
Dr. Sonnhard Graubner
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1Yes yes i got it now. So we get the final answer as (25-11x)/ (5+x)(5-x) right ? – S. Fuard Jul 09 '19 at 14:48
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For $x\ne\pm 5$, we have:
$$ \frac{8}{x+5} - \frac{3}{5-x}= $$
$$ =\frac{8(5-x)-3(5+x)}{(5+x)(5-x)}= $$
$$=\frac{40-8x-15-3x}{(5+x)(5-x)}=$$
$$=\frac{25-11x}{25-x^2}=$$
$$=\frac{11x-25}{x^2-25}.$$
MattAllegro
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