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I am a student and I am having difficulty with answering this question. I keep getting the answer wrong. Please may I have a step by step solution to this question so that I won't have difficulties with answering these type of questions in the future.

Write as a single fraction in its simplest Form.

$$\frac{3}{2x+1}+\frac{8}{2x^2-7x-4}$$

I factorised both fractions:

$$\frac{3}{(2x+1)} \qquad \frac{8}{(2x+1)(x+4)}$$ And got rid of $(2x+1)$. I don't know what to do next.

Kind Regards

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    What do mean by "got rid of $(2x+1)$"? and your factorization is wrong... – 2017-01-25
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    I was trying to simplify the expression by factorising the second fraction and found 2x+1 as the common denominator so I cancelled it out. – 2017-01-25
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    OK... what do you mean by "cancelled it out"? It can't be removed from the result. – 2017-01-25
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    I don't know how to do this type of question – 2017-01-25
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    Thank you for everyone who commented and answered. I understand how to do the question now. – 2017-01-25

3 Answers 3

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The standard approach as I understand it is:

$$\begin{align} \frac{3}{2x+1}&+\frac{8}{2x^2-7x-4} \\ &=\frac{3}{2x+1}+\frac{8}{(2x+1)(x-4)} \tag{factorise denom.}\\ &= \frac{3(x-4)}{(2x+1)(x-4)}+\frac{8}{(2x+1)(x-4)} \tag{common denom.}\\ &= \frac{3(x-4)+8}{(2x+1)(x-4)} \tag{common frac.}\\ &= \frac{3x-12+8}{(2x+1)(x-4)} \tag{multiply out}\\ &= \frac{3x-4}{(2x+1)(x-4)} \tag{result}\\ \end{align}$$

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We can take $\frac {1}{2x+1}$ common from the two terms to get $$P =\frac {3}{2x+1}+\frac {8}{(2x+1)(x-4)} $$ $$=\frac {1}{2x+1}[3 +\frac {8}{x-4 }]$$ $$=\frac {1}{2x+1}[\frac {3x-12+8}{x-4}] $$ $$=\frac {1}{2x+1}[\frac {3x-4}{x-4}] $$ Hope it helps.

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    There are many wrong signs. Make them correct. – 2017-01-25
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Getting $\frac{1}{2x+1}$ common, we have

$\frac{1}{2x+1} \left[3 + \frac{8}{x-4} \right]$

$\frac{1}{2x+1} \left[\frac{3(x-4) + 8}{x-4}\right]$

$\frac{1}{2x+1} \left[\frac{3x - 12 + 8}{x-4}\right]$

$\frac{1}{2x+1} . \frac{3x - 4}{x-4}$