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
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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)