0
$\begingroup$

I have what appears to be a simple question but am lost as how to start it. I have been asked to show that:

$$\left|\frac{64}{3x-5}-4\right|=\left|\frac{12}{3x-5}\right|\cdot|x-7|$$

Is there some sort of logical process that I can follow in this instance - a process I could put into code for a computer to follow, or do I simply need to have some sort of insight to 'see' what I need to do, because that is something I am really bad at.

Could someone please instruct me on how I should start this, and also let me know what the significance of the absolute brackets are? I find them confusing and don't understand their purpose in this question.

Thank you

$\endgroup$
  • $\begingroup$ It would definitely help if you knew about some of the properties of the absolute brackets i.e. |(x+3)(x+2)|=|x+3||x+2|, but other than that all you had to do was find a common denominator, factor the top and use the property I gave you above. $\endgroup$ – RonaldB Aug 3 '16 at 7:03
6
$\begingroup$

We have

\begin{eqnarray} \left|\frac{64}{3x-5}-4\right|&=\left|\frac{64-4(3x-5)}{3x-5}\right|\\ &=\left|\frac{84-12x}{3x-5}\right|\\ &=\left|\frac{12(7-x)}{3x-5}\right|\\ &=\left|\frac{12}{3x-5}\right||x-7|. \end{eqnarray}

$\endgroup$
  • $\begingroup$ Can you please explain what happened in the very first step? I don't understand how the 4 moved up and where the 3x-5 came from $\endgroup$ – user88720 Aug 3 '16 at 7:03
  • $\begingroup$ @user88720 I just used took the LCM of the two terms. $\endgroup$ – Karthik Aug 3 '16 at 7:05
  • $\begingroup$ Sorry I'm still doing very basic maths. I still don't follow what you have done there. You said you took the lowest common denominator of both terms (3x-5). But why did you do that? And why have you only multiplied the right side by it? Also, why did that move the 4 from the denominator to the numerator? $\endgroup$ – user88720 Aug 3 '16 at 7:11
  • $\begingroup$ @user88720 I used the following: Suppose we have $\frac{a}{b}-c$, then we can write it as $\frac{a}{b}-\frac{cb}{b}=\frac{a-cb}{b}$. Apply this with $a=64$, $b=3x-5$, and $c=4$. $\endgroup$ – Karthik Aug 3 '16 at 7:13
  • $\begingroup$ Ah righto that makes much more sense now. Thank you $\endgroup$ – user88720 Aug 3 '16 at 7:14

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.