Answer:
x can have any value, there are infinite solutions.
Step-by-step explanation:
[tex]3\left(2x-4\right)=6\left(x-2\right)[/tex]
Expand:
[tex]3\times2x-3\times 4=6x-6\times 2[/tex]
[tex]6x-12=6x-12[/tex]
Add 12 to both sides:
[tex]6x-12+12=6x-12+12[/tex]
[tex]6x=6x[/tex]
Subtract 6x from both sides:
[tex]6x-6x=6x-6x[/tex]
[tex]0=0[/tex]
There are infinite solutions for x, as no matter what the value of x is, it will not affect the end result of [tex]0=0[/tex] which is true.
