Solve SA=2A+Ph

$$S A = 2 A + P h$$

Solve for A

$$\left\{\begin{matrix}A=\frac{Ph}{S-2}\text{, }&S\neq 2\\A\in \mathrm{R}\text{, }&\left(P=0\text{ or }h=0\right)\text{ and }S=2\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 2A from both sides.

$$SA-2A=Ph$$

Combine all terms containing A.

$$\left(S-2\right)A=Ph$$

Divide both sides by S-2.

$$\frac{\left(S-2\right)A}{S-2}=\frac{Ph}{S-2}$$

Dividing by S-2 undoes the multiplication by S-2.

$$A=\frac{Ph}{S-2}$$

Solve for P

$$\left\{\begin{matrix}P=\frac{A\left(S-2\right)}{h}\text{, }&h\neq 0\\P\in \mathrm{R}\text{, }&\left(A=0\text{ or }S=2\right)\text{ and }h=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$2A+Ph=SA$$

Subtract 2A from both sides.

$$Ph=SA-2A$$

The equation is in standard form.

$$hP=AS-2A$$

Divide both sides by h.

$$\frac{hP}{h}=\frac{A\left(S-2\right)}{h}$$

Dividing by h undoes the multiplication by h.

$$P=\frac{A\left(S-2\right)}{h}$$