Dtyjlui

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!

Note that $|12-5x|=begin{cases}12-5x,&12-5xge0\5x-12,&12-5x<0end{cases}$

When $12-5xge0$, you get $12-5x=3-2ximplies x=3,12-5x=-3<0$, which is inconsistent with the initial assumption that $12-5xge0$.

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

You can divide the study into two cases:

Case 1

begin{cases}
12-5xge0 \[4px]
12-5x=-2x+3
end{cases}
that becomes
begin{cases}
xle 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 $yge0$. 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: $$|x|=xtext{ if }xgeq0text{ and }|x|=-xtext{ if }xleq0$$

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

• 
  $x=y$ if $xgeq0$
  


• 
  $-x=y$ if $xleq0$
  

Applying that correctly on the problem you mention gives:

• 
  $12-5x=-2x+3$ if $12-5xgeq0$
  


• 
  $5x-12=-2x+3$ if $12-5xleq0$
  

The first gives at first had solution $x=3$ but it must rejected because $12-5cdot3ngeq0$.

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

The final conclusion is that there are no solutions.

Note that for an absolute value function, you have:

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

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

In $vert 12-5xvert = -2x+3$, 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:

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

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

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