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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 :)

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  • $\begingroup$ Do you mean 2(x-1)(2x-1) on the second denominator? $\endgroup$ – Barbosa Sep 5 '16 at 18:16
  • $\begingroup$ Your use of parentheses is inconsistent. Is the denominator of the second fraction supposed to be $2(x - 1)(2x - 1)$? $\endgroup$ – N. F. Taussig Sep 5 '16 at 18:17
  • $\begingroup$ No, that's the expanded form $\endgroup$ – Zochonis Sep 5 '16 at 18:19
  • $\begingroup$ Take a look at the second denominator in the first line. Is there a parenthesis missing? $\endgroup$ – N. F. Taussig Sep 5 '16 at 18:21
  • $\begingroup$ Oh, yeah fixed it $\endgroup$ – Zochonis Sep 5 '16 at 18:22
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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 factor, 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, $$ 2(4x^2-3x-1)=2(x-1)(x+1/4)=1/2(x-1)(4x+1) $$

Finally $$ \frac{2(4x^2-3x-1)}{2(x-1)(2x-1)} = \frac{1/2(x-1)(4x+1)}{2(x-1)(2x-1)} = \frac{(4x+1)}{4(2x-1)} $$

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  • $\begingroup$ Thanks for the help but i don't think that's exactly what am looking for, it isn't in its simplest form $\endgroup$ – Zochonis Sep 5 '16 at 18:36
  • $\begingroup$ I put the final details in there. Make sure to understand the steps. $\endgroup$ – Barbosa Sep 5 '16 at 18:47
  • $\begingroup$ Ah thanks so much! but one more problem i have is how did you end up with 2(4x^2 - 3x - 1) when simplifying? $\endgroup$ – Zochonis Sep 5 '16 at 19:07
  • $\begingroup$ You have to do the multiplication between the expressions in parenthesis: (4x−1)(2x−1)−3 = 8x^2 -4x - 2x +1 -3 = 8x^2 - 6x - 2 = 2(4x^2 - 3x - 1) $\endgroup$ – Barbosa Sep 5 '16 at 20:26
  • $\begingroup$ Oh ok, sorry for this late comment but why did you add (2x - 1) to the first fraction at the start? $\endgroup$ – Zochonis Sep 8 '16 at 0:18

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