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


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

  • $\begingroup$ 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

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}

  • $\begingroup$ 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

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