1. x=3y-2/4,what is the value of y in terms of x ?

2. x=y^3/3y^2+y/2,what is the value of y in terms of x ?

1. I think you meant:

x = (3y-2)/4

if so, then
.... times 4
4x = 3y - 2
4x + 2 = 3y
y = (4x+2)/3

2. I will take your question the way you typed it except I think your denominator for the first fraction is probably 3y^2, so it needs brackets

x = (1/3) y + y/2
times 6
6x = 2y + 3y
6x = 5y
y = 6x/5

thank you, Reiny

To find the value of y in terms of x for each equation, we'll solve the equations for y.

1. x = 3y - 2/4

First, let's simplify the equation by multiplying both sides by 4 to get rid of the fraction:

4x = 12y - 2

Next, let's add 2 to both sides to isolate the term with y:

4x + 2 = 12y

Now, divide both sides by 12 to solve for y:

(4x + 2) / 12 = y

So, the value of y in terms of x for the first equation is:

y = (4x + 2) / 12

----------------------------------------------

2. x = y^3 / (3y^2 + y/2)

To solve this equation for y, let's begin by multiplying both sides by (3y^2 + y/2):

x * (3y^2 + y/2) = y^3

Next, distribute x to each term inside the brackets:

3xy^2 + xy/2 = y^3

Now, let's bring all y terms to one side and all x terms to the other side:

3xy^2 - y^3 = - xy/2

To get rid of the fraction, let's multiply both sides by 2:

(2)(3xy^2 - y^3) = -(2)(xy/2)

Simplifying, we have:

6xy^2 - 2y^3 = -xy

Next, let's bring all the y terms to one side:

-2y^3 - 6xy^2 + xy = 0

Now, we can factor out y:

y(-2y^2 - 6x + x) = 0

Simplifying further, we have:

y(-2y^2 - 5x) = 0

Finally, we have two possible solutions:

1. y = 0
2. -2y^2 - 5x = 0

For the second case, let's solve for y:

-2y^2 = 5x

Divide both sides by -2:

y^2 = -5x / 2

Taking the square root of both sides:

y = ± √(-5x / 2)

Thus, the value of y in terms of x for the second equation is:

y = ± √(-5x / 2)