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!
 A: 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 $12-5x<0$.
A: 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
$$
-2x+3=-\frac{30}{7}+3=-\frac{9}{7}<0
$$
A: 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: 


*

*$x \geq 0$ and $x = y$.  

*$x \leq 0$ 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: 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.
A: 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.
