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.

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.

? How do this. please explain concept behind this math.