I need help on some algebra..fractions?

Question

The problem is..

-2x + 1/3y +4 = 0 While x=2

BUT they want me to solve by first solving for Y and then substituting the X.

So far I've got..

1/3y = 2x - 4

I just need help on how to get the fraction (1/3) to the other side so that I can solve for Y.

Ten points to the best unconfusing explanation. Thanks!

Answer 1

-2x+1/3y+4=0

1/3y=2x-4

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

y=6x-12

-2x+6x-12=0

4x=12

x=3

Answer 2

                          -2x + 1/3y +4 = 0

          

                                While x=2

Condition:

                  sove for y and insert x at the end.

Solution:

(1/3)y=4-2x

y=3(4-2x)-----------------------equati... 1

put x =2 in equation 1

y=3[4-2(2)]

y=0

Answer 3

y = 6x - 12

Answer 4

Y= 6X - 12 (multiply what you had by 3, you were almost there)

So, Y = 0

Answer 5

Multiply both sides by 3

y = 6x - 12

If you substitute a 2 for x, you'll find that y = 0

Look back at the original equation, substitute 2 for x and you can see that y must be zero there also.