Answer: a = 10
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Work Shown:
[tex]y = \frac{4x+30}{x+5}\\\\y = \frac{4x+20+10}{x+5}\\\\y = \frac{(4x+20)+10}{x+5}\\\\y = \frac{4(x+5)+10}{x+5}\\\\y = \frac{4(x+5)}{x+5}+\frac{10}{x+5}\\\\y = 4+\frac{10}{x+5}\\\\[/tex]
Therefore, a = 10
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As an alternative approach, we could work in reverse like this:
[tex]y = 4+\frac{a}{x+5}\\\\y = 4*\frac{x+5}{x+5}+\frac{a}{x+5}\\\\y = \frac{4(x+5)}{x+5}+\frac{a}{x+5}\\\\y = \frac{4(x+5)+a}{x+5}\\\\y = \frac{4x+20+a}{x+5}\\\\[/tex]
Then notice that the 20+a must be 30, so solving 20+a = 30 leads to a = 10
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As yet another alternative approach (well I suppose it's actually 2 extra approaches) we could use either polynomial long division or synthetic division. Either way, you should end up with a remainder of 10, which corresponds directly to a = 10
