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I have to solve the following inequality: $$|6x−2|≤|3x−5|$$ I do know that I have to do this first step: $$|6x−2|-|3x−5|≤ 0$$ From here I got confused what I should do with the absolute values. Could someone give me a push in the right direction?

Moo
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Nawin Narain
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4 Answers4

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Square, square, you just square! Then you get $$(6x-2)^2\le (3x-5)^2.$$ Now you may perform a transposition of one term, and factorise, then the rest is trivial -- hopefully. :)

Allawonder
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  • I have tried this but I come up with a quadratic inequality where I have to use the abc formula and I have to do everything by heart. – Nawin Narain Sep 19 '19 at 20:30
  • @NawinNarain Nah, you get $(3x+3)(9x-7)\le 0,$ or $$(x+1)(x-7/9)\le 0.$$ Use your favourite method to do this: cases, or examining the signs of the factors in the intervals $(-\infty,-1],[-1,7/9],[7/9,+\infty),$ or sketching the graph of the function on LHS. I generally prefer the last two. Indeed, for quadratic functions, I almost always use the last method -- just sketch a graph to deduce the solution! – Allawonder Sep 19 '19 at 20:46
  • @NawinNarain You can use the factorisation of the difference of two squares. And faced with a quadratic you can always complete the square as a method, rather than remembering the formula - see the answer of Andre Nicholas to this question https://math.stackexchange.com/questions/49229/why-can-all-quadratic-equations-be-solved-by-the-quadratic-formula/49243#49243 – Mark Bennet Sep 19 '19 at 20:49
  • How did you come up with (3x+3)(9x−7)? My calculation gave the quadratic inequation 27x^2+6x-21≤0 which is (3x+3)(9x−7) is there a trick for this. I can't come up with this by just completing the squares or is it just me that is being special. – Nawin Narain Sep 19 '19 at 21:05
  • @NawinNarain Oh, you needn't have expanded anything. After transposing, note that you get a difference of two squares, which we can always factorise, since $a^2-b^2=(a-b)(a+b).$ – Allawonder Sep 19 '19 at 21:24
  • Oowh I get that now. If I then have the inequality I should just solve x-1≤0 and 9x−7≤0 am I right? – Nawin Narain Sep 19 '19 at 21:32
  • @NawinNarain Not quite. If you have $ab\le 0,$ then $a\le 0$ and $b\ge 0,$ or $a\ge 0$ and $b\le 0.$ – Allawonder Sep 19 '19 at 22:08
  • hmmm I see, so to determine whether it is a≤0 and b≥0 or a≥0 and b≤0 I should sketch the graph right? Is there any way of doing this without a computer because I have to do this by heart according to my course. – Nawin Narain Sep 19 '19 at 22:19
  • @NawinNarain No. My last comment was for if you want to go by cases. To use a graph just sketch the quadratic function. It has two different real roots and the leading coefficient is positive, so it's a $\cup$-shaped graph intersecting the $x$-axis at its roots. You need nothing more than that. – Allawonder Sep 19 '19 at 22:48
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    After a lot of calculation I finally got it! I am pretty weak with maths and my teacher usually gets angry with me when I take too long. I am trying to improve my maths and I absolutely love this community helping me! Thank you for your patience. – Nawin Narain Sep 19 '19 at 23:19
  • @NawinNarain I'm probably happier than you for having been of essential use to you in your learning. Well done! :D – Allawonder Sep 20 '19 at 04:38
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Hint: Start by making a sketch. Find the points of intersection of $y=|3x-5|$ and $y=|6x-2|$ by solving $|3x-5|=|6x-2|$. Then determine the range of values of $x$ for which $|3x-5|\geq|6x-2|$ (i.e. when $y=|3x-5|$ is above $y=|6x-2|$).

A. Goodier
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  • I tried to solve |3x−5|=|6x−2| and came up with the solutions x=7/9 or x=-1 But how do I determine the range of values? – Nawin Narain Sep 19 '19 at 20:07
  • This is where the sketch would help. You will then be able to see that the range of values is $-1\leq x\leq \frac{7}{9}$, since this is when $y=|3x-5|$ is above $y=|6x-2|$. – A. Goodier Sep 19 '19 at 20:09
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Consider three intervals:

$x<\dfrac13\tag1$

$\dfrac13\le x\le\dfrac53\tag2$

$x>\dfrac53.\tag3$

In interval ($1$), $3x-5, 6x-2<0$.

In interval ($2$), $6x-2\ge0 $ but $3x-5\le0$.

In interval $(3)$, $3x-5, 6x-2>0$.

Can you take it from here?

J. W. Tanner
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You can multiply inequality by $ |6x−2|+|3x−5| $. It is always greater than $ 0 $ so solutions remain the same. Then you have a simple quadratic inequality.

ngtvx
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