Solving for a variable

4x^2+3y=y+1/3 I'm stuck thanks

If all you need to do is solve for x and y then this is what I would do:

first solving for y
4x^2+3y=y+(1/3) multiply everything by 3 to get rid of (1/3)
12x^2+9y=3y then isolate y
(12x^2)/(-6)=y

then solving for x
go back a few steps to get 12x^2=-6y (this is just rearranging the y= equation)
solve that for x and you should get x=sqrt(-y/2)

i hope this helps- it seems to simple for college algebra, but there you go.

the 4th line should have been

12x^2+9y=3y + 1

you probably wanted to solve for y, thus expressing it as a function:

6y = -12x^2 + 1
y = -2x^2 + 1/6

To solve for a variable, you typically want to isolate that variable on one side of the equation. Let's solve for the variable x in the equation 4x^2 + 3y = y + 1/3.

Step 1: Move the terms involving x to one side of the equation.
Start by subtracting y from both sides:
4x^2 = y + 1/3 - 3y

Simplified: 4x^2 = -2y + 1/3

Step 2: Isolate x by dividing both sides by 4:
x^2 = (-2y + 1/3) / 4

Step 3: Take the square root of both sides to eliminate the exponent of 2:
x = ± √[(-2y + 1/3) / 4]

Now, you have solved for the variable x in terms of y.