Question

How to solve for z in terms of x and y?
2(2)(x^2+y^2)(2x+2yz)=100[xz+y]

Answer 1

\(2(2)(x^2+y^2)(2x+2yz)=100(xz+y)\)
Is this the question?

Answer 2

yup.

Answer 3

\(4(2x^3+2xy^2+2x^2yz+2y^3z)=100(xz+y)\)
\(2x^3+2xy^2+2x^2yz+2y^3z=25xz+25y\)
\(2x^3+2xy^2-25y=25xz-2x^2yz-2y^3z\)
Factorise RHS.
Take z as the common factor from \(25xz-2x^2yz-2y^3z\)
Divide by \(25x-2x^2y-2y^3\) both sides.

Answer 4

or what u can do is,just for convenience, take
\[a=4(x^2+y^2)\]
to get
\[2ax+2ayz=100xz+100y\]
solving for z here is easier
in the end u can resubstitute the value of a.....

Answer 5

Thank you guys! I get it now:)

Answer 6

yw.:)