Absolute Value Equations

**Bashyboy** · September 3rd 2011, 08:57 AM

How, when you are given an absolute value equation, solve it graphically on a real-number line?

**Prove It** · September 3rd 2011, 09:04 AM

Originally Posted by Bashyboy

How, when you are given an absolute value equation, solve it graphically on a real-number line?

You should know that the absolute value corresponds to the "size" of a number...

So say |x| = 3. That means that the size of the number is 3 units from 0. Therefore x = -3 or x = 3. Does that make sense?

**Bashyboy** · September 3rd 2011, 09:23 AM

Yes, it does. But I become a bit more confounded with a problem like |x - 3| = 5. How would I solve one like this on a real number-line?

**Prove It** · September 3rd 2011, 09:38 AM

Originally Posted by Bashyboy

Yes, it does. But I become a bit more confounded with a problem like |x - 3| = 5. How would I solve one like this on a real number-line?

Well that means that the size of x - 3 is 5 units from 0.

So $\displaystyle x - 3 = -5 \implies x = -2$ or $\displaystyle x - 3 = 5 \implies x = 8$ .

**Bashyboy** · September 5th 2011, 10:46 AM

I have another problem. I am given a real number line with the two points -3 an 1 marked on it. How do I write an absolute value equations from this data?

**Plato** · September 5th 2011, 10:58 AM

Originally Posted by Bashyboy

I have another problem. I am given a real number line with the two points -3 an 1 marked on it. How do I write an absolute value equations from this data?

If $a<b$ then the equation $\left| {x - \frac{{b + a}}{2}} \right| = \frac{{b - a}}{2}$ has solutions $x=a\text{ or }x=b$ .

**Bashyboy** · September 5th 2011, 11:05 AM

I am truly sorry, but I don't quite follow. I have not seen anything like this before.

**Plato** · September 5th 2011, 11:08 AM

Originally Posted by Bashyboy

I am truly sorry, but I don't quite follow. I have not seen anything like this before.

Are you saying that you cannot substitute $a=-3~\&~b=1$ into that?

**Bashyboy** · September 5th 2011, 11:10 AM

No, plugging and chugging is quite simple. But I have never seen that formula you have presented me with. Also, the way the book wants me to solve it is just graphically.

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