# Inequalities, Absolute Values on both sides

• Jun 23rd 2011, 04:53 AM
Zellator
Inequalities, Absolute Values on both sides
Hi Forum,
That's just a little question I have, and I can't find the answer on the web.
Find the solution set for $|2x+1|>|x-5|$

Now, the thing I have learned is to keep the left expression (or the right, it doesn't matter) and keep and inverse the sign of the other expression, changing the direction accordingly.
Like this:

$|2x+1|>|x-5|$

$2x+1>x-5$
$2x-x>-1-5$

$x>-6$

and

$2x+1<-x+5$
$2x+x<-1+5$

$x<\frac{4}{3}$

But if we are to get the squares of both sides we get:
$4x^2+4x+1>x^2-10x+25$
$3x^2+14x-24>0$

$(3x-4)(x-6)>0$

Getting the roots here and looking at the sign of $a$ in the quadratic expression we can see that see that we are searching for x>0 values of x that will give a x<a u b<x interval. So the solution set is:

$x>-6$ U $\frac{4}{3}

Now, calculating using the other method gave us the inverse interval.
So the direction of the inequality must change when we take the absolute value out.
Is this the correct way of doing it? Why is this possible?

Thanks!
• Jun 23rd 2011, 06:02 AM
Re: Inequalities, Absolute Values on both sides
You could also consider...

$x<-\frac{1}{2}\Rightarrow\ 2x+1<0;\;\;\;x-5<0$

$-(2x+1)>-(x-5)\Rightarrow\ -2x-1>5-x\Rightarrow\ -6>x$

The "absolute value" inequality is true for x<-6

$x>-\frac{1}{2}\Rightarrow\ 2x+1>0$

$x-5>0\Rightarrow\ x>5$

Hence, between -0.5 and 5, we have $x-5<0$

Therefore

$2x+1>-(x-5)\Rightarrow\ 3x>5-1\Rightarrow\ x>\frac{4}{3}$

Beyond x=5, both are positive

$2x+1>x-5\Rightarrow\ x>-6$

which is absolutely true as x>5, so for x>5

$|2x+1|>|x-5|$

Hence

$-\infty

$\frac{4}{3}
• Jun 23rd 2011, 06:10 AM
Prove It
Re: Inequalities, Absolute Values on both sides
With two moduli, you need to consider several separate cases, noting that

$\displaystyle |2x + 1| = \begin{cases}2x+1 \textrm{ if }x \geq -\frac{1}{2} \\ -(2x + 1)\textrm{ if }x < -\frac{1}{2}\end{cases}$

and

$\displaystyle |x - 5| = \begin{cases}x - 5\textrm{ if }x \geq 5\\ -(x - 5)\textrm{ if }x < 5\end{cases}$.

So the cases you need to consider are:

Case 1:
$\displaystyle x < -\frac{1}{2}$, which would mean $\displaystyle |2x + 1| = -(2x + 1)$ and $\displaystyle |x - 5| = -(x - 5)$

Case 2:
$\displaystyle -\frac{1}{2} \leq x < 5$, which would mean $\displaystyle |2x + 1| = 2x + 1$ and $\displaystyle |x - 5| = -(x - 5)$

Case 3:
$\displaystyle x \geq 5$, which would mean $\displaystyle |2x + 1| = 2x + 1$ and $\displaystyle |x - 5| = x - 5$

So now you need to solve your original equation $\displaystyle |2x + 1| = |x - 5|$ for each of these cases. I will do the first and leave the rest for you.

Case 1:
\displaystyle \begin{align*} -(2x + 1) &= -(x - 5) \\ 2x + 1 &= x - 5 \\ x &= -6 \end{align*}

Since this is in the required domain $\displaystyle x < -\frac{1}{2}$, this answer is acceptable.

Now you try for the remaining two cases.
• Jun 23rd 2011, 06:59 AM
Plato
Re: Inequalities, Absolute Values on both sides
Quote:

Originally Posted by Zellator
Hi Forum,
That's just a little question I have, and I can't find the answer on the web.
Find the solution set for $|2x+1|>|x-5|$

the squares of both sides we get:
$4x^2+4x+1>x^2-10x+25$
$3x^2+14x-24>0$
$(3x-4)(x-6)>0$
$x{\color{red}{>}}-6$ U $\frac{4}{3}
Is this the correct way of doing it? Why is this possible?

Yes, in fact I prefer that way of doing this problem.
It is based on $|a|>|b|\text{ if and only if }a^2>b^2~.$
But you have a mistake in the answer.
It should be $x{\color{blue}{<-6}}\cup \frac{4}{3}
• Jun 23rd 2011, 07:08 AM
Re: Inequalities, Absolute Values on both sides
In fact,
the factoring is incorrect...

viz

$3x^2+14x-24>0$

$(3x-4)(x+6)>0$

Both factors are positive or both are negative.
• Jun 23rd 2011, 10:59 AM
Zellator
Re: Inequalities, Absolute Values on both sides
Quote:

Beyond x=5, both are positive

$2x+1>x-5\Rightarrow\ x>\color{blue}-6$

which is absolutely true as x>5

Quote:

Originally Posted by Prove It
Case 3:
$\displaystyle x \geq 5$, which would mean $\displaystyle |2x + 1| = 2x + 1$ and $\displaystyle |x - 5| = x - 5$

Solving the 3rd case

$2x+1=x-5$

$x=-6$

hence

$x>-6$
for $x \geq5$

Now, isn't this redundant information?
I just don't understand the relevance of this part of your explanations, since this doesn't weight in the final answer.
Why doesn't it matter? If it is there it must have some usefulness, right?

Sorry about the mistake Plato, I kept changing that so it would be better to understand. I changed it too much :)

Thanks!
• Jun 23rd 2011, 11:36 AM
Re: Inequalities, Absolute Values on both sides
Yes,

when x>5, both 2x+1 and x-5 are >0.

Then we can examine for what x, the given inequality holds.
(Just in case it did not hold for all x above 5).

2x+1>x-5, so x>-6.
This means that x has to be >-6 when x>5
and since any x>5 is also >-6,
then 2x+1>x+5 for all x>5.

It ends up quite redundant, but we needed to check, having gone that route.
The significance of writing that is...

imagine we were examining a similar inequality,
and we found that for x>5, with both expressions positive,
we found that the inequality held for x>7.
This would the mean that x would have to be >7 and >5,
hence only x>7 is both >5 and >7.

So that's the logic and Plato's preference is vindicated.
Thanks for pointing out my error too!
• Jun 23rd 2011, 12:10 PM
Zellator
Re: Inequalities, Absolute Values on both sides
OK, that's very very clear right now.
I get the redundant part too!
Thanks for going on the down the hard road on the explanation.
Most people would just point out that $a^2>b^2$ is easier and leave it like that.
I guess we learn much more from alternative methods.

Thanks again,
All the best.
• Jun 23rd 2011, 04:37 PM
HallsofIvy
Re: Inequalities, Absolute Values on both sides
Another way to solve general inequalities is to first solve the associated equation.

|2x+1|= |x- 5| means that either (1) 2x+ 1= x- 5 or (2)2x+ 1= -(x- 5)

In the first case, 2x- x= -1- 5 or x= -6.

In the second case, 2x+ 1= -x+ 5 so 2x+ x= 5- 1, 3x= 4, x= 4/3.

The point is that both |2x+ 1| and |x- 5| are continuous functions so they change from ">" to "<" and vice-versa only where they are "=".

The two points, x= -6, and x= 4/3, divide the real numbers into three intervals, x< -6, -6< x< 4/3, and x> 4/3.

-7 < -6 and if x= -7, |2x+1|= 13 while |x- 5|= 12. Yes, 13> 12. x= -7 satisfies the inequality so any x<-6 satisfies it.

-6< 0< 4/3 and if x= 0, |2x+ 1|= 1 while |x- 5|= 5. No, 1< 5. x= 0 does not satisfy the inequality so no value of x between -6 and 4/3 satisfies it.

2> 4/3 and if x= 2, |2x+1|= 5 while |x- 5|= 2. Yes, 13> 2. x= 2 satisfies the inequality so any x> 4/3 satisfies it.
• Jun 24th 2011, 02:36 AM
Zellator
Re: Inequalities, Absolute Values on both sides
Quote:

Originally Posted by HallsofIvy
Another way to solve general inequalities is to first solve the associated equation.

|2x+1|= |x- 5| means that either (1) 2x+ 1= x- 5 or (2)2x+ 1= -(x- 5)

In the first case, 2x- x= -1- 5 or x= -6.

In the second case, 2x+ 1= -x+ 5 so 2x+ x= 5- 1, 3x= 4, x= 4/3.

The point is that both |2x+ 1| and |x- 5| are continuous functions so they change from ">" to "<" and vice-versa only where they are "=".

The two points, x= -6, and x= 4/3, divide the real numbers into three intervals, x< -6, -6< x< 4/3, and x> 4/3.

-7 < -6 and if x= -7, |2x+1|= 13 while |x- 5|= 12. Yes, 13> 12. x= -7 satisfies the inequality so any x<-6 satisfies it.

-6< 0< 4/3 and if x= 0, |2x+ 1|= 1 while |x- 5|= 5. No, 1< 5. x= 0 does not satisfy the inequality so no value of x between -6 and 4/3 satisfies it.

2> 4/3 and if x= 2, |2x+1|= 5 while |x- 5|= 2. Yes, 13> 2. x= 2 satisfies the inequality so any x> 4/3 satisfies it.

Hi HallsOfIvy!
This is where I was trying to get when I first came upon this question.
It really seems more natural to solve the equation, the way you did it.
Thanks!
That's a great way of solving it.
• Jun 24th 2011, 03:50 AM
Re: Inequalities, Absolute Values on both sides
Another way to consider this is...

As the graphs of the functions are linear, their graphs are V-shaped.
There are 3 cases.

(1) The graphs of the functions touch the x-axis at the same point and have different slopes.
Then the function with the steeper slope lies above the other except at the point of intersection on the x-axis.

(2) The graphs touch the x-axis at different points and have the same slopes.
Then one is greater than the other at only one side of the single point of intersection.
(We have identical V's crossing at a single point).

(3) The V's have different positive slopes (and similarly different negative slopes)
and touch the x-axis at seperate points.
Then the V with steeper slope lies above the other to the left and right of the 2 points of intersection.