# Interesting Absolute Value problem

1. Jul 3, 2011

### flyingpig

1. The problem statement, all variables and given/known data

Find all x in

|5x-2| = 6x-12

3. The attempt at a solution

http://www.wolframalpha.com/input/?i=Plot[|5x-2|+%3D+6x-12%2C{x%2C-1.5%2C1.5}]

Notice how only x = 10 is a solution?

So my question is, why isn't x = -14/11 a solution? I mean sure they don't intersect on the graph, but is there anyway of inspection on the equation to know that x = -14/11 is not a solution?

2. Jul 3, 2011

### Angelrel

Hi, the solutions are:

5x-2=6x-12 x=10

2-5x=6x-12 x=14/11

why do you say -14/11?

3. Jul 3, 2011

### Mentallic

Because while we are solving $5x-2=6x-12$ and $5x-2=-(6x-12)$ we also have to place restrictions of the values of x in each of these cases.

For example, in the first case we are assuming $5x-2\geq 0$ which means $x\geq 2/5$ and now if we get a value of $x<2/5$ then we have a contradiction and that solution is not valid.
For the second case where we assume $5x-2<0$ thus $x<2/5$ but we instead get an answer of $x=14/11>2/5$ this answer is invalid, so we scrap it.

4. Jul 4, 2011

### AC130Nav

But, if instead, we make fewer assumptions and take just the original formula at face value
|5x-2| = 6x-12 obviously means $0\leq 6x-12$ by the definition of absolute values. This means $x\geq 2$. This eliminates the same false answer for the OP.

Is this incorrect? And what relation do the values involving 2/5 have?

5. Jul 4, 2011

### SammyS

Staff Emeritus
Look at this graph from http://www.wolframalpha.com/input/?i=plot+|5x-2|+,+6x-12,+x=-4..12".

A purely algebraic approach:
$$|5x-2|=\left\{\begin{array} , 5x-2&,&\text{if }\ 5x-2\ge0\\ -5x+2&,&\text{if }\ 5x-2<0\end{array}\right.\quad=\left\{\begin{array} , 5x-2&,&\text{if }\ x\ge\frac{2}{5}\\ -5x+2&,&\text{if }\ x<\frac{2}{5}\end{array}\right.$$
Case1: x ≥ 2/5, therefore |5x-2| = 5x-2

Solving |5x-2| = 6x-12 gives 5x-2 = 6x-12 which gives x = 10 .

Case2: x<2/5, therefore |5x-2| = -5x+2

Solving |5x-2| = 6x-12 gives -5x+2 = 6x-12 which gives x = 14/11 .
but 14/11 > 2/5, so it is inconsistent with this case.

Therefore, there is only one solution, x = 10 .

Last edited by a moderator: Apr 26, 2017
6. Jul 5, 2011

### Mentallic

No, it's correct
The method SammyS and I used by breaking it into cases and testing each one has a correlation with solving inequalities involving fractions where the unknown variable is in the denominator.

Yours is much simpler though

7. Jul 12, 2011

### flyingpig

You are my hero.

Last edited by a moderator: Apr 26, 2017