# Write as a single fraction in its simplest form

So it seems as though am having quite a bit of trouble with this question of mine, i've spent some time trying to find the correct way to do it but am unsure about how to do it since i always seem to end up in a slight muddle i guess.

The question:

Write as a single fraction in its simplest form:

$\frac{4x-1}{2(x-1)}$ - $\frac{3}{2(x-1)(2x-1)}$

I've tried expanding the brackets first to end up with:

$\frac{4x-1}{2x-2}$ - $\frac{3}{2x-2(2x-1)}$

(All of the signs are negative if you couldn't tell)

And i was then going to combine each of the denominators with their respective numerators (opposite method) and then simplify and go from there...

I've tried different methods too but i dohn't think any of the ones i have tried are accurate and am not entirely sure if this one is either :/

A detailed method would be greatly appreciated but ill be happy if any advice is provided.

Thank a bunch :)

• Do you mean 2(x-1)(2x-1) on the second denominator? Commented Sep 5, 2016 at 18:16
• Your use of parentheses is inconsistent. Is the denominator of the second fraction supposed to be $2(x - 1)(2x - 1)$? Commented Sep 5, 2016 at 18:17
• No, that's the expanded form Commented Sep 5, 2016 at 18:19
• Take a look at the second denominator in the first line. Is there a parenthesis missing? Commented Sep 5, 2016 at 18:21
• Oh, yeah fixed it Commented Sep 5, 2016 at 18:22

Anytime you want to operate fractions like that you need them to have the same denominator. You can get that easily if the denominator is presented as the product of its factors, like in your problem.

So $$\frac{4x-1}{2(x-1)} - \frac{3}{2(x-1)(2x-1)}$$ is almost ready for the subtraction, all we need is for the first fraction to have a $$(2x-1)$$ on the denominator, just like the second fraction. But we can get that if we multiple the first fraction by this missing factor like this $$\frac{(4x-1)(2x-1)}{2(x-1)(2x-1)} - \frac{3}{2(x-1)(2x-1)}$$ (multiplying the numerator and denominator of a fraction by the same number doesn't change its value since it is the same as multiplying by one, $$\frac{2x-1}{2x-1}=1$$).

Now we are ready to operate with the fractions $$\frac{(4x-1)(2x-1)}{2(x-1)(2x-1)} - \frac{3}{2(x-1)(2x-1)} = \frac{(4x-1)(2x-1)-3}{2(x-1)(2x-1)}.$$

Simplifying $$\frac{(4x-1)(2x-1)-3}{2(x-1)(2x-1)} = \frac{2(4x^2-3x-1)}{2(x-1)(2x-1)}.$$

The solutions for $$4x^2-3x-1=0$$ are $$x=1$$ and $$x=-\frac{1}{4}$$. Thus, $$2(4x^2-3x-1)=2\cdot 4(x-1)(x+1/4)=2(x-1)(4x+1).$$

Finally $$\frac{2(4x^2-3x-1)}{2(x-1)(2x-1)} = \frac{2(x-1)(4x+1)}{2(x-1)(2x-1)} = \frac{4x+1}{2x-1},$$ with $$x \neq 1$$.

• Thanks for the help but i don't think that's exactly what am looking for, it isn't in its simplest form Commented Sep 5, 2016 at 18:36
• I put the final details in there. Make sure to understand the steps. Commented Sep 5, 2016 at 18:47
• Ah thanks so much! but one more problem i have is how did you end up with 2(4x^2 - 3x - 1) when simplifying? Commented Sep 5, 2016 at 19:07
• Thanks @RyanG for pointing it out. I believe that it should be correct now. Commented Aug 15, 2021 at 16:54
• No worries. The condition $x\neq1$ that is implicit in the original expression still needs to be retained in final, simplified expression (by making it explicit in the last line). Commented Aug 15, 2021 at 17:03

This may be a bit late now but to anyone looking at the question now. The answer is actually $$(4x+1)/(2x-1)$$. The original guy who answered made a mistake with the factorisation of $$(4x^2-3x-1)$$.

Just use symbolab or something.