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

Question

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!

Answer 1

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 $12-5x<0$.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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 $$ -2x+3=-\frac{30}{7}+3=-\frac{9}{7}<0 $$