Absolute Value Inequalities

[axrw]

This seems like it should be simple, but naturally I'm confused. How would you go about solving absolute value inequalities like:

1) |3x - 1| > 5x - 2

2) |x + 2| <= |x - 5|

3) |x - 3| + |2x + 5| > 6

4) |p - 4| + |p + 4| < 8

On the first inequality I know the answer is x < 1/2, but I keep getting x < 3/8 too.

3x - 1 < -5x + 2 or 3x - 1 > 5x - 2

and from those I get

x < 3/8 or x < 1/2

What am I doing wrong?

And I don't even know where to begin on the other ones.

 

[jonboy]

Hello axrw!

I'll try to be a help.

|3x - 1| > 5x - 2

Since we have a >, you must solve: 3x - 1 > 5x - 2 or 3x - 1 < -5x + 2

|x + 2| <= |x - 5|

I think this one is all real numbers.

|x - 3| + |2x + 5| > 6

Combine like terms: |3x + 2| > 6

Once again, this is a >. So solve: 3x + 2 > 6 or 3x + 2 < -6

|p - 4| + |p + 4| < 8

Combine like terms: |2p| < 8

Since we have a <, you need to solve: -8 < 2p < 8

 

[pka]

|x + 2| <= |x - 5|

I think this one is all real numbers.

Is it true for \(\displaystyle x=6\)? No it is not.

Solve this \(\displaystyle \left( {x + 2} \right)^2 \le \left( {x - 5} \right)^2\).

 

[stapel]

1) |3x - 1| > 5x - 2

Try doing cases. If 3x - 1 > 0 (that is, if x > 1/3), then |3x - 1| = 3x - 1. So you have two cases: x > 1/3 and x < 1/3. Deal with each separately:

. . .x > 1/3:

. . . . .|3x - 1| = 3x - 1 > 5x - 2
. . . . .-1 > 2x - 2
. . . . .1 > 2x
. . . . .1/2 > x
. . . . .x < 1/2

Since this case depends on x > 1/3, you then have 1/3 < x < 1/2 for this case. Now do the other one:

. . .x < 1/3:
. . . . .|3x - 1| = -(3x - 1) > 5x - 2
. . . . .-3x + 1 > 5x - 2
. . . . .1 > 8x - 2
. . . . .3 > 8x
. . . . .3/8 > x
. . . . .x < 3/8

But 3/8 = 9/24 and 1/3 = 8/24. Since x is already less than 1/3 = 8/24, then the actual solution for this case is x < 1/3.

Combining the two solutions, you get "x < 1/2". Note that this matches what you got, and can be confirmed by graphing y₁ = abs(3x - 1) and y₂ = 5x - 2 in your calculator. The absolute-value function is greater than (higher than) the linear function for all x < 1/2.

2) |x + 2| <= |x - 5|

You have more cases here: x + 2 > 0 for x > -2, and x - 5 > 0 for x > 5. So the intervals are x < -2, -2 < x < 5, x > 5. Work with each separately.

3) |x - 3| + |2x + 5| > 6

Uses cases again: x < -5/2, -5/2 < x < 3, and x > 3.

4) |p - 4| + |p + 4| < 8

This one works just like (3).

Eliz.

 

[axrw]

Thank you for the reply. I'm gonna try the second.

So on the second the intervals are (-inf, -2)U[-2, 5)U[5,inf)

So the first interval both of the expressions inside the bars are negative:

-(x + 2) <= -(x - 5)

working out leads to

0 <= 7

Which means the set of all real numbers, right? But the interval was for all numbers below -2, so that isn't a solution. So it's just x < -2

The second interval

x + 2 <= -(x - 5)

leads to

x <= 3/2

Which is in the interval and is a solution. So it's -2 <= x <= 3/2

On the third interval both are positive

x + 2 <= x - 5

and I got

0 <= -7, which just doesn't make sense at all, so do I throw out the [5, inf) interval altogether?

So combining them all I end up with x <= 3/2?

Sorry to be such a bother, but I'm still not sure whether I'm getting this or not. I really appreciate the help though.

 

[pka]

In many cases it is much easier to use this method.
\(\displaystyle \begin{array}{rcl}
\left| a \right| \le \left| b \right|\quad \mbox{if and only if} \quad a^2 \le b^2 & & \\
\left| {x + 2} \right| & \le & \left| {x - 5} \right| \\
\left( {x + 2} \right)^2 & \le & \left( {x - 5} \right)^2 \\
x^2 + 4x + 4 & \le & x^2 - 10x + 25 \\
14x & \le & 21 \\
x & \le & \frac{3}{2} \\
\end{array}\)

 

[axrw]

Ok, yeah, that is easier. Don't suppose there is a trick for ones like |a| + |b| < c?

Thanks for the tip pka.

 

[jonboy]

Thanks for the tip pka.

I double that.