Multiple modulus inequalities : GMAT Quantitative Section
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# Multiple modulus inequalities

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19 Dec 2009, 01:01
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Hi guys

Need to know how to solve multiple modulus inequalities

Ex - Solve |x-10| - |2x+3| <= |5x-10|

I do understand that we need to first find the roots, which happen to be 10, -3/2 and 2 in this case

we can then divide them into intervals -

x<-3/2
-3/2<x<2
2<x<10
x>10

what do we do post this division into intervals?
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19 Dec 2009, 02:47
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zaarathelab wrote:
Hi guys

Need to know how to solve multiple modulus inequalities

Ex - Solve |x-10| - |2x+3| <= |5x-10|

I do understand that we need to first find the roots, which happen to be 10, -3/2 and 2 in this case

we can then divide them into intervals -

x<-3/2
-3/2<x<2
2<x<10
x>10

what do we do post this division into intervals?

If we dot he way you are proposing:

Yes, there are 3 check point at which absolute values in the example change their sign: -3/2, 2, and 10 (te values of x for which the expression in || equals to zero). Then you should expand the modulus in these ranges:

A. $$x<-\frac{3}{2}$$ --> $$|x-10|-|2x+3|\leq{|5x-10|}$$ --> $$-(x-10)+(2x+3)\leq{-(5x-10)}$$ --> $$6x\leq{-3}$$ --> $$x\leq{-\frac{1}{2}$$ --> as we considering the range when $$x<-\frac{3}{2}$$, then finally we should write $$x<-\frac{3}{2}$$:
----($$-\frac{3}{2}$$)---------------------------------

B. $$-\frac{3}{2}\leq{x}\leq{2}$$ --> $$-(x-10)-(2x+3)\leq{-(5x10)}$$ --> $$2x<=3$$ --> $$x<={\frac{3}{2}$$ --> $${-\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}$$;
----($$-\frac{3}{2}$$)----($$\frac{3}{2}$$)-------------------------

C. $$2<x<10$$ --> $$-(x-10)-(2x+3)\leq{5x-10}$$ --> $$17\leq{8x}$$ --> $$x\geq{\frac{17}{8}}$$ --> $${\frac{17}{8}}\leq{x}<10$$;
---------------------($$\frac{17}{8}$$)--------($$10$$)----

D. $$x\geq{10}$$ --> $$(x-10)-(2x+3)\leq{5x-10}$$ --> $$-3\leq{6x$$ --> $$x\geq{-\frac{1}{2}}$$ --> $$x\geq{10}$$;
---------------------($$\frac{17}{8}$$)--------($$10$$)----

After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are:
$$x\leq{\frac{3}{2}}$$ and $${\frac{17}{8}}\leq{x}$$
----($$-\frac{3}{2}$$)----($$\frac{3}{2}$$)----($$\frac{17}{8}$$)--------($$10$$)-----

One important detail when you define the critical points (10, -3/2 and 2) you should include them in either of intervals you check. So when you wrote:
x<-3/2
-3/2<x<2
2<x<10
x>10

You should put equal sign for each of the critical point in either interval. For example:
x=<-3/2
-3/2<x<=2
2<x<1=0
x>10

Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality.
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01 Feb 2012, 01:31
Could somebody explain this in detail, please?

I didn't get the concept of roots here, it doesn't make the equation zero - isn't it?
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01 Feb 2012, 01:47
bsaikrishna wrote:
Could somebody explain this in detail, please?

I didn't get the concept of roots here, it doesn't make the equation zero - isn't it?

When x<=0 then |x|=x and if x>0 then |x|=-x.

For |x-10|: when x<=10 then |x-10|=-(x-10) (because if x<=10 then x-10<=0) and when x>10 then |x-10|=x-10 (because if x>10 then x-10>0). So, zaarathelab calls "roots" (you can also call "critical point" or "check point") the values of x for which the expression in || equals to zero: in two different cases when x<that value and x>that value the absolute values expands with different sign (it switches the sign at this value).

Check Absolute Values chapter of Math Book for more: math-absolute-value-modulus-86462.html

Hope it helps.
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01 Feb 2012, 01:55
Hi,

I solved the modulus by taking the positive & negative scenarios of each modulus. I solved them separately & arrived at 1/2 <= x & x<= -1/2.

Am I on the right track or messed up some where??
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Last edited by boomtangboy on 01 Feb 2012, 02:05, edited 1 time in total.
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01 Feb 2012, 01:59
boomtangboy wrote:
Hi,

I solved the modulus by taking the positive & negative scenarios of each modulus. I solved them separately & arrived at 1/2 <= x & x<= -1/2.

Am I on the right track or messed up some where??

The answer to the initial question is x<=3/2 or 17/8<=x, so "1/2 <= x & x<= -1/2" is not right. Check my previous post: you should check different cases for different ranges to get the correct answer.
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04 Oct 2012, 10:45
Bunuel wrote:
zaarathelab wrote:
Hi guys

Need to know how to solve multiple modulus inequalities

Ex - Solve |x-10| - |2x+3| <= |5x-10|

I do understand that we need to first find the roots, which happen to be 10, -3/2 and 2 in this case

we can then divide them into intervals -

x<-3/2
-3/2<x<2
2<x<10
x>10

what do we do post this division into intervals?

If we dot he way you are proposing:

Yes, there are 3 check point at which absolute values in the example change their sign: -3/2, 2, and 10 (te values of x for which the expression in || equals to zero). Then you should expand the modulus in these ranges:

A. $$x<-\frac{3}{2}$$ --> $$|x-10|-|2x+3|\leq{|5x-10|}$$ --> $$-(x-10)+(2x+3)\leq{-(5x-10)}$$ --> $$6x\leq{-3}$$ --> $$x\leq{-\frac{1}{2}$$ --> as we considering the range when $$x<-\frac{3}{2}$$, then finally we should write $$x<-\frac{3}{2}$$:
----($$-\frac{3}{2}$$)---------------------------------

B. $$-\frac{3}{2}\leq{x}\leq{2}$$ --> $$-(x-10)-(2x+3)\leq{-(5x10)}$$ --> $$2x<=3$$ --> $$x<={\frac{3}{2}$$ --> $${-\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}$$;
----($$-\frac{3}{2}$$)----($$\frac{3}{2}$$)-------------------------

C. $$2<x<10$$ --> $$-(x-10)-(2x+3)\leq{5x-10}$$ --> $$17\leq{8x}$$ --> $$x\geq{\frac{17}{8}}$$ --> $${\frac{17}{8}}\leq{x}<10$$;
---------------------($$\frac{17}{8}$$)--------($$10$$)----

D. $$x\geq{10}$$ --> $$(x-10)-(2x+3)\leq{5x-10}$$ --> $$-3\leq{6x$$ --> $$x\geq{-\frac{1}{2}}$$ --> $$x\geq{10}$$;
---------------------($$\frac{17}{8}$$)--------($$10$$)----

After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are:
$$x\leq{\frac{3}{2}}$$ and $${\frac{17}{8}}\leq{x}$$
----($$-\frac{3}{2}$$)----($$\frac{3}{2}$$)----($$\frac{17}{8}$$)--------($$10$$)-----

One important detail when you define the critical points (10, -3/2 and 2) you should include them in either of intervals you check. So when you wrote:
x<-3/2
-3/2<x<2
2<x<10
x>10

You should put equal sign for each of the critical point in either interval. For example:
x=<-3/2
-3/2<x<=2
2<x<1=0
x>10

Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality.

Hi Brunel,

How do we decide the signs in front of the modulus in the equation.
Eg:: |x-10| - |2x+3| <= |5x-10|......say we are looking in an interval - 3/2 to 2. then based on what concentration do we put +or - sign in front of the modulus.

and please explain >> step 1----------x = sqrt (x^2) ?
< step w--------------- => x = |x| ? How do this. please explain concept behind this math.

Thanks!!
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Math Expert
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05 Oct 2012, 03:50
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Maverick04308 wrote:
Hi Brunel,

How do we decide the signs in front of the modulus in the equation.
Eg:: |x-10| - |2x+3| <= |5x-10|......say we are looking in an interval - 3/2 to 2. then based on what concentration do we put +or - sign in front of the modulus.

and please explain >> step 1----------x = sqrt (x^2) ?
< step w--------------- => x = |x| ? How do this. please explain concept behind this math.

Thanks!!

Explained here: multiple-modulus-inequalities-88174.html#p1037473

Hope it helps.
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Re: Multiple modulus inequalities   [#permalink] 10 Sep 2015, 01:31
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# Multiple modulus inequalities

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