# Why am I getting two solutions for this absolute value equation?

The question is "State with a reason whether there are any solutions to |12-5x| = -2x+3"

I can clearly see there are no solutions when I graph it but I've learned to solve these questions doing the following:

$$|x| = y$$

$$x = y$$

$$x = -y$$

When doing this here, I get:

$$|12 - 5x| = -2x + 3$$

$$12 - 5x = -2x + 3$$

$$12 - 5x = 2x - 3$$

Solving for each of these I get $$(3, -3)$$ -> So no solution here as the y is negative - makes sense

But I also get $$(15/7, 9/7)$$ which would, in theory be an intersection.

Obviously this isn't right but algebraically I'm having trouble with the intuition.

Hope someone can help!

• Welcome to the website. Please use Mathjax to typeset your equations for better presentation. In this case, you need only enclose your equations by the \$ symbol. – Shubham Johri Jan 10 at 16:16

Note that $$|12-5x|=\begin{cases}12-5x,&12-5x\ge0\\5x-12,&12-5x<0\end{cases}$$

When $$12-5x\ge0$$, you get $$12-5x=3-2x\implies x=3,12-5x=-3<0$$, which is inconsistent with the initial assumption that $$12-5x\ge0$$.

When $$12-5x<0$$, you get $$12-5x=2x-3\implies x=15/7,12-5x=9/7>0$$, which is inconsistent with the initial assumption that $$12-5x<0$$.

You can divide the study into two cases:

## Case 1

$$\begin{cases} 12-5x\ge0 \\[4px] 12-5x=-2x+3 \end{cases}$$ that becomes $$\begin{cases} x\le 12/5 \\[4px] x=3 \end{cases}$$ No solution.

## Case 2

$$\begin{cases} 12-5x<0 \\[4px] 5x-12=-2x+3 \end{cases}$$ that becomes $$\begin{cases} x>12/5 \\[4px] x=15/7 \end{cases}$$ No solution.

## Where did you go wrong?

In order that $$|x|=y$$ holds, it's necessary that $$y\ge0$$. For $$x=15/7$$, you have $$-2x+3=-\frac{30}{7}+3=-\frac{9}{7}<0$$

So, what you've learned is... I'm not going to say wrong, exactly, but more "only occasionally right". More specifically, $$|x| = y$$ if and only if one of the following holds:

1. $$x \geq 0$$ and $$x = y$$.
2. $$x \leq 0$$ and $$x = -y$$.

You have considered only the second half of each of these.

Now, let's solve your question: if $$12 - 5x > 0$$, then we require $$12 - 5x = -2x + 3$$. Rearranging that, we have $$x = 3$$. But with $$x = 3$$, we have $$12 - 5x = -3 < 0$$, so we are not in this case, and this is not a solution.

On the other hand, if $$12 - 5x < 0$$, then we require $$12 - 5x = 2x - 3$$, but then $$x = \frac{9}{7}$$, and $$12 - 5x = \frac{39}{7} > 0$$, so again, we are not in this case, and this is not a solution.

Thus, we have no solutions.

A correct start would be: $$|x|=x\text{ if }x\geq0\text{ and }|x|=-x\text{ if }x\leq0$$

Then equation $$|x|=y$$ splits up in two cases that must be discerned:

• $$x=y$$ if $$x\geq0$$
• $$-x=y$$ if $$x\leq0$$

Applying that correctly on the problem you mention gives:

• $$12-5x=-2x+3$$ if $$12-5x\geq0$$
• $$5x-12=-2x+3$$ if $$12-5x\leq0$$

The first gives at first had solution $$x=3$$ but it must rejected because $$12-5\cdot3\ngeq0$$.

The second gives at first had solution $$x=\frac{15}7$$ but it must rejected because $$12-5\cdot\frac{15}7\nleq0$$.

The final conclusion is that there are no solutions.

Note that for an absolute value function, you have:

$$f(x) = \vert x\vert = \begin{cases} x; \quad x \geq 0 \\ -x; \quad x < 0 \end{cases}$$

You haven’t put any emphasis on the conditions (that $$\vert x\vert = x$$ if $$x \geq 0$$ and that $$\vert x\vert = -x$$ if $$x < 0$$), which are extremely important.

In $$\vert 12-5x\vert = -2x+3$$, you make two distinct cases, but you must also remember the constraints:

$$\begin{cases} \ 12-5x = -2x+3 ; \quad \color{blue}{12-5x \geq 0} \\ 12-5x = -(-2x+3) = 4x-3; \quad \color{blue}{12-5x < 0} \end{cases}$$

For case one, ignoring the extra condition, you get $$x = 3$$. Plug it in the condition given and see if it is met:

$$12-5x \overset{?}{\geq} 0; \quad x = 3$$

$$12-5(3) = 12-15 = -3 \ngeq 0$$

This is false, and is also reflected when the RHS becomes negative, as you correctly spotted.

For case two, ignoring the condition, you get $$x = \frac{15}{7}$$. Plug it in the condition given and see if it is met:

$$12-5x \overset{?}{<} 0; \quad x = \frac{15}{7}$$

$$12-5\left(\frac{15}{7}\right) = 12-\frac{75}{7} = \frac{9}{7} \nless 0$$

This is also false, so the equation has no solution.