Consider this system of equations:

2x+1/4y=3 (equation A)

2/3x-y=6 (equation B)

The expressions that give the value of y are___ and ___

The solution for the given system is___?

You need to rewrite the equation as it relates to y. (Isolate y)

Equation A 2x + 1/4 y = 3

Step 1 Subtract 2x from both sides 2x-2x+1/4y=3+2x. Since 2x-2x = 0 we now have 1/4 y = 3 + 2x.

Step 2 Divide all numbers by 1/4 so we will have y.

1/4y divided by1/4 = 3 divided by1/4 + 2x divided 1/4. Since 1/4 divided by 1/4 is 1 we end up with

y = 12 + 8x Note (3 divded by 1/4 is 12 and 2 divided by 1/4 is 8)

Equation B

2/3x - y = 6.

Step 1. Add y to both sides so we can get a poitive y

2/3x - y + y = 6 + y or 2/3x = 6 + y.

Now subtract 6 from both sides to isolate y

2/3x - 6 = 6 - 6 + y. and we get

2/3x - 6 = y.

To find the solution we use y = y and substitue the equations

So 2/3x - 6 = 12 + 8x

Now solve for x 2/3x - 2/3 x - 6 = 12 + 8x - 2/3x or - 6 = 12 + 7 1/3x.

Then - 6 - 12 = 12 - 12 + 7 1/3x or - 18 = 7 1/3 x

Now divide bith sides by 7 1/3 to solve for x then plug back into one of the equations to find y.