Remember that , so I would rewrite this equation as
.
Solve for .
and 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).
Therefore the lines do not intersect above the x-axis
and are therefore not a solution of
although the value of x is a solution to the equation.
so these lines also do not intersect above the x-axis.
therefore these lines do intersect above the x-axis.
therefore these lines also intersect above the x-axis.
Hence the solution is the union of
and
That's a "cute" calculation you made!
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......
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
....No, since the minimum value of 3x is 9.
Case 2:
Case 3: Both negative within modulus