You could also consider...
The "absolute value" inequality is true for x<-6
Hence, between -0.5 and 5, we have
Therefore
Beyond x=5, both are positive
which is absolutely true as x>5, so for x>5
Hence
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
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:
and
But if we are to get the squares of both sides we get:
Getting the roots here and looking at the sign of 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:
U
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!
You could also consider...
The "absolute value" inequality is true for x<-6
Hence, between -0.5 and 5, we have
Therefore
Beyond x=5, both are positive
which is absolutely true as x>5, so for x>5
Hence
With two moduli, you need to consider several separate cases, noting that
and
.
So the cases you need to consider are:
Case 1:
, which would mean and
Case 2:
, which would mean and
Case 3:
, which would mean and
So now you need to solve your original equation for each of these cases. I will do the first and leave the rest for you.
Case 1:
Since this is in the required domain , this answer is acceptable.
Now you try for the remaining two cases.
About this 3rd case, both Archie Meade and Prove It talked about it.
Solving the 3rd case
hence
for
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
Sorry about the error in the factoring Archie Meade
Thanks!
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!
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 is easier and leave it like that.
I guess we learn much more from alternative methods.
Thanks again,
All the best.
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.
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.