Solve G=MF-5/PD^3-3A

$$G = \frac { M F - 5 } { P D ^ { 3 } } - 3 A$$

Solve for A

$$A=\frac{FM-GPD^{3}-5}{3PD^{3}}$$

$$P\neq 0\text{ and }D\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by PD^{3}.

$$GPD^{3}=MF-5-3APD^{3}$$

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

$$MF-5-3APD^{3}=GPD^{3}$$

Subtract MF from both sides.

$$-5-3APD^{3}=GPD^{3}-MF$$

Add 5 to both sides.

$$-3APD^{3}=GPD^{3}-MF+5$$

The equation is in standard form.

$$\left(-3PD^{3}\right)A=5+GPD^{3}-FM$$

Divide both sides by -3PD^{3}.

$$\frac{\left(-3PD^{3}\right)A}{-3PD^{3}}=\frac{5+GPD^{3}-FM}{-3PD^{3}}$$

Dividing by -3PD^{3} undoes the multiplication by -3PD^{3}.

$$A=\frac{5+GPD^{3}-FM}{-3PD^{3}}$$

Divide GD^{3}P-MF+5 by -3PD^{3}.

$$A=-\frac{5+GPD^{3}-FM}{3PD^{3}}$$

Solve for D

$$\left\{\begin{matrix}D=\sqrt[3]{\frac{FM-5}{P\left(3A+G\right)}}\text{, }&\left(M\neq \frac{5}{F}\text{ or }F=0\right)\text{ and }G\neq -3A\text{ and }P\neq 0\\D\neq 0\text{, }&G=-3A\text{ and }M=\frac{5}{F}\text{ and }F\neq 0\text{ and }P\neq 0\end{matrix}\right.$$