Solving Abs Value Inequalities

[Iceycold12]

Just need to make sure my notes are correct.

We know that if we get |-3x| = 18 we split that into -3x = 18 or -3x = -18. Then solve and get two solutions.

However, if we get say 5|y+3|<15 (Note it's not an equal sign now but an inequality sign). We divide by 5 to get abs value alone, so y+3 < 15 or y + 3 >-15. Now that's where I get confused. Do we solve it by solving these two: y+3<15 or y+3>-15 and getting two solutions or putting it in this form: -15 < y+3 < 15 which solved equals -> -18<y<12. The original equations is in |x|< a form. The solved form of -a < x < a I got form purple math:

This pattern for "less than" absolute-value inequalities always holds: Given the inequality | x | < a, the solution is always of the form –a < x < a. Even when the exercises get more complicated, the pattern still holds. Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

• Solve | 2x + 3 | < 6.Since this is a "less than" absolute-value inequality, the first step is to clear the absolute value according to the pattern. Then I'll solve the linear inequality.
  • | 2x + 3 | < 6
    –6 < 2x + 3 < 6 [this is the pattern for "less than"]
    –6 – 3 < 2x + 3 – 3 < 6 – 3
    –9 < 2x < 3
    ^–9/₂ < x < ³/₂
  Then the solution to | 2x + 3 | < 6 is the interval ^–9/₂ < x < ³/₂.

If I need to solve an abs value inequality in |x| > a form, I split it into two, like it says here:

The pattern for "greater than" absolute-value inequalities always holds: the solution is always in two parts. Given the inequality | x | > a, the solution always starts by splitting the inequality into two pieces:x < –a or x > a. For instance:

• Solve | 2x – 3 | > 5.The first thing to do is clear the absolute value bars by splitting the inequality into two. Then I'll solve the two linear inequalities.
  • | 2x – 3 | > 5
    2x – 3 < –5 or 2x – 3 > 5 [this is the pattern for "greater than"]
    2x < –2 or 2x > 8
    x < –1 or x > 4
  This PAIR of inequalities is the solution;
  the solution to | 2x – 3 | > 5 consists of the two intervals x < –1 and x > 4.

Basically, the point of my post is to make sure I've taken my notes correctly: For abs value equations |x| = a I answer it with "or" with 2 solutions. For abs value inequalities in |x| < a form, I answer in -a < x < a form.
And lastly in |x| > a form, I answer with 2 solutions using "or" the same as abs value equations basically.

 

[pka]

That is basically correct.
If \(\displaystyle c>0\):
\(\displaystyle |x-a|\le c\) has solutions \(\displaystyle a-c\le x\le a+c\),

\(\displaystyle |x-b|\ge c\) has solutions \(\displaystyle x\ge b+c\) or \(\displaystyle x\le b-c~.\)

 

[Iceycold12]

Thanks pka, but about the second equation you showed using or:

Alrighty so when solving abs value, problems with = sign and |x| > a form have both "or" answers?

 

[pka]

Thanks pka, but about the second equation you showed using or:
Alrighty so when solving abs value, problems with = sign and |x| > a form have both "or" answers?

Example
\(\displaystyle |x+1|=2\) has solutions \(\displaystyle x=1\text{ OR }x=-3\)

\(\displaystyle |x+1|\ge 2\) has solutions \(\displaystyle x\ge 1\text{ OR }x\le -3\).
In set notation that is \(\displaystyle (-\infty,-3]\cup [1,\infty)\).

 

[Iceycold12]

Got it, thanks a lot.

Little side question, say a problem is |x| = -a, then it splits to |x| = -a or |x| = -a right? A is already negative. After about 30 or so problems I got a scenario like this. Usually a is positive then it splits to a positive a and -a. Not sure what happens here.

 

[HallsofIvy]

Got it, thanks a lot.

Little side question, say a problem is |x| = -a, then it splits to |x| = -a or |x| = -a right?

No, for several reasons. If |x|= a, where a is positive, then x= a or x= -a- there is no | |. Also whether you mean "|x|= -a or |x|= -a" or "x= -a or x=-a", those are the same! Finally, if |x|= -a, where a is negative, then u= -a is positive so |x|= -a= u leads to "either x= u= -a or x= -u= -(-a)= a".

A is already negative. After about 30 or so problems I got a scenario like this. Usually a is positive then it splits to a positive a and -a. Not sure what happens here.

Becareful about inequalities. You can think of |x- a| as meaning the distance between x and a. So if |x- 5|> 4, x can be any point whose distance from 5 is greater than 4. 5+ 4= 9 and 5- 4= 1 have distance from 5 equal to 4 so point farther than that are x< 1 and x> 5, two disjoint sets. But |x- 5|< 4 means x is closer to 5 than 4, or closer to 5 than 1 or 9. x must satisfy x> 1 and (not "or") x< 9, the single interval (1, 9).

 

[JeffM]

If we get say 5|y+3|<15. We divide by 5 to get abs value alone, so y+3 < 15 or y + 3 >-15. NO!!!

Just as with equations, you must perform identical operations on both sides of the inequality. As for your other question, you have a choice:

\(\displaystyle 5|y + 3| < 15 \implies 0 \le |y + 3| < 3 \implies 0 \le y + 3 < 3\ OR\ - 3 < (y + 3) \le 0 \implies -3 \le y < 0\ OR\ -6 < y \le -3 \implies\)

\(\displaystyle -6 < y < 0.\)

You can simplify this: \(\displaystyle 5|y + 3| < 15 \implies |y + 3| < 3 \implies -3 < y + 3 < 3 \implies -6 < y < 0.\)

The two ways come out to the same thing when |a| < b.

Basically, the point of my post is to make sure I've taken my notes correctly: For abs value equations |x| = a I answer it with "or" with 2 solutions. For abs value inequalities in |x| < a form, I answer in -a < x < a form.
And lastly in |x| > a form, I answer with 2 solutions using "or" the same as abs value equations basically. You have got it.

.