please check my work if i done correctly. I'm little confused.
Remember that $\displaystyle \displaystyle |X| = \sqrt{X^2}$, so I would rewrite this equation as
$\displaystyle \displaystyle \sqrt{(2x-6)^2} = \sqrt{(4-5x)^2}$
$\displaystyle \displaystyle (2x- 6)^2 = (4 - 5x)^2$.
Solve for $\displaystyle \displaystyle x$.
$\displaystyle 2x-6$ and $\displaystyle 4-5x$ are both linear functions.
One has a positive slope and the other has a negative slope, so the lines intersect at a single point.
The modulus of these are V-shaped as we invert the part of the graph below the x-axis,
hence there are now two points of intersection if the original lines do not intersect on the x-axis.
Hence, we can look for the points of intersection of 4 lines, while eliminating points of intersection below the y-axis
(refer to the dashed lines on the attachment).
$\displaystyle 2x-6=4-5x\Rightarrow\ 7x=10\Rightarrow\ x=\frac{10}{7}$
$\displaystyle 2\left(\frac{10}{7}\right)-6<0$
Therefore the lines do not intersect above the x-axis
and are therefore not a solution of
$\displaystyle |2x-6|=|4-5x|$
although the value of x is a solution to the equation.
$\displaystyle 2x-6=-(4-5x)\Rightarrow\ 2x-6=5x-4\Rightarrow\ -2=3x\Rightarrow\ x=-\frac{2}{3}$
$\displaystyle 2\left(-\frac{2}{3}\right)-6<0$
so these lines also do not intersect above the x-axis.
$\displaystyle -(2x-6)=4-5x\Rightarrow\ 6-2x=4-5x\Rightarrow\ 3x=-2\Rightarrow\ x=-\frac{2}{3}$
$\displaystyle -\left[2\left(-\frac{2}{3}\right)-6\right]>0$
therefore these lines do intersect above the x-axis.
$\displaystyle -(2x-6)=-(4-5x)\Rightarrow\ 6-2x=5x-4\Rightarrow\ 7x=10\Rightarrow\ x=\frac{10}{7}$
$\displaystyle -\left[2\left(\frac{10}{7}\right)-6\right]>0$
therefore these lines also intersect above the x-axis.
Hence the solution is the union of
$\displaystyle -(2x-6)=(4-5x)$
and
$\displaystyle -(2x-6)=-(4-5x)$
That's a "cute" calculation you made!
$\displaystyle (2x-6)^2-(4-5x)^2=0\Rightarrow\ [(2x-6)+(4-5x)][(2x-6)-(4-5x)]=0$
Yes, it's a good shortcut and locates the values of x that are the solutions of the modulus function.
This is because the x co-ordinates of the points of intersection of the lines
are the same as the points of intersection of the parabolas.
The method I showed earlier will give the x and y co-ordinates if required,
however only the x co-ordinates are requested.
Without reference to graphs, you can also try this way......
$\displaystyle |2x-6|=|5x-4|$
When x>3, both 2x-6 and 5x-4 are both positive
and when 5x<4, both 2x-6 and 5x-4 are both negative.
Case 1: both positive within modulus
$\displaystyle x\ge3,\;\;5x\ge4\Rightarrow\ x\ge3$
$\displaystyle (2x-6)=(5x-4),\;\;\;x\ge3$
$\displaystyle -2=3x\;\;?\;\;\;x\ge3$ ....No, since the minimum value of 3x is 9.
Case 2:
$\displaystyle 2x-6\le0,\;\;5x-4\ge0\;\;\;\Rightarrow\ 5x\ge4,\;\;x\le3$
$\displaystyle -(2x-6)=5x-4,\;\;\;\frac{4}{5}\le x\le3$
$\displaystyle 6-2x=5x-4\Rightarrow\ 10=7x\;\;?\;\;\;\frac{4}{5}\le x\le3$
$\displaystyle x=\frac{10}{7}$
Case 3: Both negative within modulus
$\displaystyle x\le\frac{4}{5}$
$\displaystyle -(2x-6)=-(5x-4)\Rightarrow\ 6-2x=4-5x\Rightarrow\ 3x=-2\;\;?\;\;\;x\le\frac{4}{5}$
$\displaystyle x=-\frac{2}{3}$