Mathematics OpenStudy (anonymous):

Solving absolute value Inequalities with more than one absolute value term: Say for example, I have: 2|x - 4| < |2x - 4| (this was a question I made up!) How would I go about with solving this? OpenStudy (anonymous):

you have to break it down in to cases and it is a pain OpenStudy (anonymous):

I know how to do with one absolute value argument. OpenStudy (anonymous):

Ok wait so I will post, can you confirm if they are correct? OpenStudy (anonymous):

sure because the one you posted looks goofy OpenStudy (anonymous):

2|x - 4| < |2x - 4| SHOOT Im stuck. OpenStudy (anonymous):

BEcause OpenStudy (anonymous):

WAit OpenStudy (anonymous):

ok we can do this OpenStudy (anonymous):

Could we work with how I used to do it with one argument? This is how it went. OpenStudy (anonymous):

Say I had the following: |3 - 2x| ≤ 5 OpenStudy (anonymous):

We would do it like this: Case 1: 3 - 2x ≥ 0 AND 3 - 2x ≤ 5 or Case 2: 3 - 2x < 0 AND -(3-2x) ≤ 5 OpenStudy (anonymous):

So how would we go about when there are two arguments? OpenStudy (anonymous):

hold on these are different OpenStudy (anonymous):

and harder! OpenStudy (anonymous):

2|x - 4| < |2x - 4| Ok so how would we do it, using the definition of absolute value? OpenStudy (anonymous):

$|3-2x|<5$ easy algebra, we say $-5<2x-3<5$ $-8<2x<2$ $-4<x<1$ OpenStudy (anonymous):

Yea OpenStudy (anonymous):

but for this one, unless you want to graph (which may be easier) you have to break it in to cases OpenStudy (anonymous):

Now how are we suppose to solve this bad boy. OpenStudy (anonymous):

Ok let us go the case way. OpenStudy (anonymous):

ok $|2x-4| = \left\{\begin{array}{rcc} 2x-4 & \text{if} & x \geq 2 \\ 4-2x& \text{if} & x < 2 \end{array} \right.$ OpenStudy (anonymous):

How did you type that OpenStudy (anonymous):

and $2|x-4| = \left\{\begin{array}{rcc} 2x-8 & \text{if} & x \geq 4 \\ 8-2x& \text{if} & x < 4 \end{array} \right.$ OpenStudy (anonymous): OpenStudy (anonymous):

Wait, I am confused by how you got 4, and 2 for the "ifs" OpenStudy (anonymous):

oh that is easy enough, $|2x-4|=2x-4 \text{ if }2x-4>0\iff 2x>4\iff x>2$ etc OpenStudy (anonymous):

Got it. OpenStudy (anonymous):

ok so now we have to check over the intervals. suppose $x>4$ then you have $2x-8<2x-4$ which is true for all x, so we know this works if $x>4$ OpenStudy (anonymous):

So then by the definition of absolute value, |x - 4| = (x-4), if x ≥ 4 -(x-4), if x < 4 OpenStudy (anonymous):

exactly OpenStudy (anonymous):

Ok now what are we suppose to do with these defined cases? OpenStudy (anonymous):

so we see that this inequality is satisfied for all x > 4 right? because if x is greater than 4 you are comparing $2x-8$ with $2x-4$ and it is always less OpenStudy (anonymous):

Give me one minute to go check above. OpenStudy (anonymous):

ok i wrote it out take your time OpenStudy (anonymous):

Wait how could you go x > 4? Is it not x ≥ 4? OpenStudy (anonymous):

ok fine be picky lol OpenStudy (anonymous):

You said we have to check the intervals, suppose "x > 4", but in the definition, we did x ≥ 4 OpenStudy (anonymous):

Ok could you resay what you did with x ≥ 4? OpenStudy (anonymous):

sure and i will write it slowly OpenStudy (anonymous):

if $x\geq 4$ then $|2x-4|=2x-4$ and $2|x-4|=2x-8$ so the inequality is $2x-8\leq 2x-4$ $-8\leq-4$ which is always true. true for all x OpenStudy (anonymous):

now we have to check on the interval $2<x<4$ OpenStudy (anonymous):

So basically, you took these intervals and created a number line? OpenStudy (anonymous):

< -------------- 0 ---- 2 ---- 4 ------------------> OpenStudy (anonymous):

think i messed that up. on this interval $2|x-4|=8-2x$ and $|2x-4|=2x-4$ right? OpenStudy (anonymous):

yes exactly OpenStudy (anonymous):

so now we have to solve $8-2x<2x-4$ OpenStudy (anonymous):

Ok so you are saying we have to use the intervals that can be created from: < -------------- 0 ---- 2 ---- 4 ------------------> and test each one to see if they satisfy the inequalities OpenStudy (anonymous):

So in this case, we would have x < 2, 2 < x < 4, and x > 4 OpenStudy (anonymous):

$12<4x$ $3<x$ so now we know if works if x > 3 OpenStudy (anonymous):

So in this case, we would have x < 2, 2 < x < 4, and x > 4 ? OpenStudy (anonymous):

right you have to check three cases you have it! OpenStudy (anonymous):

YES!!!!!!!!!!!!!!!!!!!!!11 OpenStudy (anonymous):

in the middle we see it works if $x>3$ OpenStudy (anonymous):

Ok so this is the basic idea: you get the definition, and you get the numbers for the number line after the "if" statement OpenStudy (anonymous):

Ok let me try OpenStudy (anonymous):

and we already know it works if $x>4$ so now we know it works if $x>3$ OpenStudy (anonymous): OpenStudy (anonymous):

Ok say we do x < 2, then |x-4| will became -(x-4) and |2x -4| will became -(2x - 4) 2-(x-4) < -(2x - 4) 2(-x + 4) < (-2x + 4) -2x + 8 < -2x + 4 8 < 4 This is not true. OpenStudy (anonymous):

right! OpenStudy (anonymous):

so putting it all together, the right hand piece always works, the middle tells you x > 3, and the left hand piece never works, so solution is $x>3$ OpenStudy (anonymous):

now this one was probably easier to graph because both of these are piecewise lines, and you could just graph them and see where one was above the other. but this is the algebra way to do it. OpenStudy (anonymous):

Would you mind showing me how you would graph it? OpenStudy (anonymous):

here is a picture, you will have to copy and past because the link won't work http://www.wolframalpha.com/input/?i=2 |x-4|%3C|2x-4| OpenStudy (anonymous):

the absolute value signs mess up the link, so copy and paste rather than clicking OpenStudy (anonymous):

So the basic idea is this: you get the numbers that follow after the "if" in the definition, put them on the number line and establish intervals to test out with the original inequality. If it works out, then it is a solution? OpenStudy (anonymous):

So in this case we got 2, and 4, so if I put them on the number line, I would get: x < 2, 2 < x < 4 and x > 4. OpenStudy (anonymous):

right. you have to use the correct definition of the function in each case OpenStudy (anonymous):

absolute value always a pain. it looks so easy when you first see it because it just says "make it positive" but it is a piecewise function so you constantly have to consider what interval you are working over. OpenStudy (anonymous):

you got this now right? OpenStudy (anonymous):

that didn't work OpenStudy (anonymous):

if you want to see how i wrote a piecewise function, right click on the expression and select "show source" it will give it to you

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