Solve P(0)quadP(y)=P(y)P(1)+9

$$P ( 0 ) \quad P ( y ) = P ( y ) P ( 1 ) + 9$$

Solve for y

$$y=-\frac{9}{P^{2}}$$

$$P\neq 0$$

Steps for Solving Linear Equation

Multiply P and P to get P^{2}.

$$P^{2}\times 0y=PyP\times 1+9$$

Multiply P and P to get P^{2}.

$$P^{2}\times 0y=P^{2}y\times 1+9$$

Anything times zero gives zero.

$$0=P^{2}y\times 1+9$$

Swap sides so that all variable terms are on the left hand side.

$$P^{2}y\times 1+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$P^{2}y\times 1=-9$$

Reorder the terms.

$$yP^{2}=-9$$

The equation is in standard form.

$$P^{2}y=-9$$

Divide both sides by P^{2}.

$$\frac{P^{2}y}{P^{2}}=-\frac{9}{P^{2}}$$

Dividing by P^{2} undoes the multiplication by P^{2}.

$$y=-\frac{9}{P^{2}}$$

Solve for P (complex solution)

$$P=-3iy^{-\frac{1}{2}}$$

$$P=3iy^{-\frac{1}{2}}\text{, }y\neq 0$$

Solve for P

$$P=3\sqrt{-\frac{1}{y}}$$

$$P=-3\sqrt{-\frac{1}{y}}\text{, }y<0$$