3/4-1/3(1/2y-2)=3(y-1/4) ??? and how?

3/4-1/3(1/2y-2)=3(y-1/4)

$$\\\frac{3}{4}-\frac{1}{3}(\frac{1}{2}y-2)=3(y-\frac{1}{4})\\\\ \frac{3}{4}-\frac{1}{3}(\frac{y}{2}-2)=3(y-\frac{1}{4})\\\\ \frac{3}{4}-\frac{y}{6}+\frac{2}{3}=3y-\frac{3}{4}\\\\ $Now it will be easier if there were no fractions$\\ $The lowest common multiple of 4,6 and 3 is 12$\\\\ $So multiply both sides by 12 to get rid of all the fractions$\\ 12(\frac{3}{4}-\frac{y}{6}+\frac{2}{3})=12(3y-\frac{3}{4})\\\\ \frac{12*3}{4}-\frac{12y}{6}+\frac{12*2}{3}=12*3y-\frac{12*3}{4}\\\\ 9-2y+8=36y-9\\\\ 17=38y-9\\\\ 26=38y\\\\$$

$$\\y=\frac{26}{38}\\\\ y=\frac{13}{19}\\\\$$

thank you so much that really helped me out, i cant tell you how much clearer it is now

whats it called when you multiplyed it by 12? is there a term for it?

Thank you, I am glad I helped you.

Well, you are multiplying by the lowest common denominator.