# Thread: Simple absolute value Q

1. ## Simple absolute value Q

The question:
Solve for x,
|2x-1|+|x+3| = 5

What's the best way to attempt questions involving the addition of absolute values? I can consider the case when x is negative, and x is positive, but that seems like an elementary approach. Is there a better way?

Note: I'm not interested in the solution, I'm interested in the method. Thanks.

2. I suppose you could always graph the function, see where it crosses the line $\displaystyle \displaystyle y = 5$...

3. Why would considering the two cases where x is negative and positive be an elementary approach? Also, even if it is, what's wrong with an elementary approach? By doing this, we get x = 1 and -1. I don't see anything wrong with that. As Prove It has already said, you could graph it, but that seems like more work than is necessary.

4. Originally Posted by rtblue
Why would considering the two cases where x is negative and positive be an elementary approach? Also, even if it is, what's wrong with an elementary approach? By doing this, we get x = 1 and -1. I don't see anything wrong with that. As Prove It has already said, you could graph it, but that seems like more work than is necessary.
I just figured there'd be a better way. For instance, with this example:

|2x + 1| = x + 1

You could square both sides and apply the quadratic formula to get a solution, instead of looking at negative/positive. Just wondering if there was a similar method.

5. In this case you would probably have to square both sides twice...

$\displaystyle \displaystyle |2x - 1| + |x + 3| = 5$

$\displaystyle \displaystyle (|2x-1| + |x+3|)^2 = 5^2$

$\displaystyle \displaystyle (2x - 1)^2 + 2|2x-1||x+3| + (x + 3)^2 = 25$

$\displaystyle \displaystyle 2|2x-1||x+3| = 25 - (2x-1)^2 - (x+3)^2$.

Squaring both sides again will eliminate the absolute values, but you'll end up with a quartic you have to solve...

6. Indeed.

With the graphing method, is there a trick to doing it, or is it a matter of subbing in values to determine the shape?

7. Draw the graphs of $\displaystyle \displaystyle |2x - 1|$ and $\displaystyle \displaystyle |x + 3|$, then use addition of ordinates to graph $\displaystyle \displaystyle |2x - 1| + |x + 3|$.

8. I expected that'd be the case. Thanks again.

9. Algebrically, you have to take three cases so that absolute values are eliminated.
1)x<=-3
2)-3<x<0.5
3) x>=0.5

and consider the accepted solutions

10. Originally Posted by Glitch
The question:
Solve for x,
|2x-1|+|x+3| = 5

What's the best way to attempt questions involving the addition of absolute values? I can consider the case when x is negative, and x is positive, but that seems like an elementary approach. Is there a better way?

Note: I'm not interested in the solution, I'm interested in the method. Thanks.
Similarly, you could say

Both $\displaystyle 2x-1$ and $\displaystyle x+3$ are $\displaystyle \ge\ 0$ gives

$\displaystyle |2x-1|+|x+3|=2x-1+x+3=5$

and for that case you must have $\displaystyle 2x\ge\ 1$ and $\displaystyle x\ge\ -3$ so $\displaystyle 2x\ge\ 1$

Next, $\displaystyle 2x-1\ge\ 0$ and $\displaystyle x+3\le\ 0$

This is not possible as we cannot have $\displaystyle 2x-1\ge\ 0$ and $\displaystyle x\le\ -3$

However, we can have $\displaystyle 2x-1\le\ 0$ and $\displaystyle x\ge\ -3$

In this case you have $\displaystyle (1-2x)+x+3=5$

(You must verify that the solution given falls correctly within the required range)

Finally, if $\displaystyle 2x-1$ and $\displaystyle x+3$ are both $\displaystyle \le\ 0$

you find that $\displaystyle (1-2x)-(x+3)=5$ requires x to be above $\displaystyle -3$
which causes a contradiction.

So two of the 4 options are valid.