Question

(p+q)(z-px-qy)=1

Answer 1

$F=pz-p^2x-pqy+qz-pqx-q^2y-1=0$

$\frac{dx}{F_p}=\frac{dy}{F_q}=\frac{dz}{pF_p+qF_q}=-\frac{dp}{F_x+pF_z}=-\frac{dq}{F_y+qF_z}$

$\frac{dx}{z-2px-qy-qx}=\frac{dy}{-py+z-px-2qy}=\frac{dz}{p(z-2px-qy-qx)+q(-py+z-px-2qy)}=$

$=-\frac{dp}{-p^2-pq+p(p+q)}=-\frac{dq}{-pq-q^2+q(p+q)}$

$\frac{dp}{-p^2-pq+p(p+q)}=\frac{dp}{0}\implies p=a=const$

$\frac{dq}{-pq-q^2+q(p+q)}=\frac{dq}{0}\implies q=b=const$

Substitute values into the initial equatio.

Then:

$(a+b)(z-ax-by)=1$

$z-ax-by=c_1$

$\frac{dx}{z-2px-qy-qx}=\frac{dy}{-py+z-px-2qy}=\frac{dp}{0}$

$dy-dx=dp=a$

$y-x=c_2$

The general solution:

$F(c_1,c_2)=F(z-ax-by, y-x)=0$