Answer to Question #294781 in Algebra for Prithika

Question

Solve (y ^ 2 * z)/x * p + xzq = y ^ 2 .

Answer 1

$y^2zp$ $+x^2zq=y^2x$

The subsidiary equations are

$\frac{dx}{y^2z}=\frac{dy}{x^2z}=\frac{dz}{y^2x}$

Considering the first two terms of the equation we get

$\frac{dx}{y^2}=\frac{dy}{x^2}$

$x^2dx-y^2dy=0$

$x^3-y^3=c_1$

Considering the 1^st and last terms of the equation

$\frac{dx}{y^2z}=\frac{dz}{y^2x}$

$\frac{dx}{z}=\frac{dz}{x}$

$xdx-zdz=0$

$x^2-z^2=c_2$

Solution

$f(x^3-y^3,x^2-z^2)=0$