how would you solve for y in this problem:
x=(y^2+3y)^(1/3)
would it equal: y= x^3/y - 3 ?
and what about this problem:
x= 1/[(y^2+3y-5)^3]
would it equal: y = y/x + 5/y - 3?
I would appreciate your help very much!
<<how would you solve for y in this problem:
x=(y^2+3y)^(1/3)
would it equal: y= x^3/y - 3 ? >>
Your equation is correct but it does not allow you to solve explicitly for y. What you have to do is treat the "x-containing" term as if it were a constant in the quadratic equation:
y^2 + 3y -x^3 = 0
Then use the equation for the root of quadratic equations:
y = [-6 +/- sqrt (9 + 4x^3)]/2
The second question can be handled similarly.
For the second problem:
<<x= 1/[(y^2+3y-5)^3]
would it equal: y = y/x + 5/y - 3? >>
To solve for y, start by isolating the denominator on one side:
1/x = (y^2 + 3y - 5)^3
Next, take the cube root of both sides to eliminate the exponent:
(1/x)^(1/3) = y^2 + 3y - 5
Now, rearrange the equation to have y terms on the left side:
y^2 + 3y - (1/x)^(1/3) - 5 = 0
This is a quadratic equation in terms of y. We can solve it using the quadratic formula:
y = (-3 +/- sqrt(9 - 4(1)(-((1/x)^(1/3)) - 5)))/(2(1))
Simplifying further would be quite complex given the presence of the expression (1/x)^(1/3). Therefore, the solution for y is given by the equation above, but it can't be simplified further.
I hope this helps! Let me know if there's anything else I can assist you with.