1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ What do mean by "got rid of $(2x+1)$"? and your factorization is wrong... $\endgroup$ – Joffan Jan 25 '17 at 16:55
  • $\begingroup$ 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. $\endgroup$ – OliviaAages Jan 25 '17 at 16:57
  • $\begingroup$ OK... what do you mean by "cancelled it out"? It can't be removed from the result. $\endgroup$ – Joffan Jan 25 '17 at 17:02
  • $\begingroup$ I don't know how to do this type of question $\endgroup$ – OliviaAages Jan 25 '17 at 17:03
  • $\begingroup$ Thank you for everyone who commented and answered. I understand how to do the question now. $\endgroup$ – OliviaAages Jan 25 '17 at 17:24
1
$\begingroup$

The standard approach as I understand it is:

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

$\endgroup$
0
$\begingroup$

Getting $\frac{1}{2x+1}$ common, we have

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

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

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

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ There are many wrong signs. Make them correct. $\endgroup$ – Kanwaljit Singh Jan 25 '17 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.