Kelly is quite right that the denominator can not be zero. So let's not forget about that as we go through our steps.
\(\rm \frac12+\frac3x=\frac34\)
To get rid of the fractions let's multiply both sides of the equation by the Least Common Multiple of our denominators. We see a 2, 4 and x. The LCM of these values is 4x so we'll multiply both sides by 4x.
\(\rm 4x\left(\frac12+\frac3x\right)=\left(\frac34\right)4x\)
distribute the 4x to each term on the left,
and cancel stuff out as needed,
and do the same on the right side of the equation,
\(\rm 2x+12=3x\)
We've taken x out of the denominator though. So we should keep a note somewhere on the side of the page that x can not equal zero still. Even though we moved some things around, make our equation look different, we still have that restriction.
\(\rm 2x+12=3x\qquad\qquad\qquad x
e0\)
From this point, solve for x using Algebra:
Subtract 2x from each side,
\(\rm 12=x\)
12 is a solution, and it is not 0.
If the solution had worked out to be x=0, we would reject it.
