Solve TC(q)=q^3/3+q^2+4q+100

$$T C ( q ) = \frac { q ^ { 3 } } { 3 } + q ^ { 2 } + 4 q + 100$$

Solve for C

$$\left\{\begin{matrix}C=\frac{q^{3}+3q^{2}+12q+300}{3Tq}\text{, }&q\neq 0\text{ and }T\neq 0\\C\in \mathrm{R}\text{, }&q=\sqrt[3]{2\sqrt{5263}-145}-\sqrt[3]{2\sqrt{5263}+145}-1\text{ and }T=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 3.

$$3TCq=q^{3}+3q^{2}+12q+300$$

The equation is in standard form.

$$3TqC=q^{3}+3q^{2}+12q+300$$

Divide both sides by 3Tq.

$$\frac{3TqC}{3Tq}=\frac{q^{3}+3q^{2}+12q+300}{3Tq}$$

Dividing by 3Tq undoes the multiplication by 3Tq.

$$C=\frac{q^{3}+3q^{2}+12q+300}{3Tq}$$

Solve for T

$$\left\{\begin{matrix}T=\frac{q^{3}+3q^{2}+12q+300}{3Cq}\text{, }&q\neq 0\text{ and }C\neq 0\\T\in \mathrm{R}\text{, }&q=\sqrt[3]{2\sqrt{5263}-145}-\sqrt[3]{2\sqrt{5263}+145}-1\text{ and }C=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 3.

$$3TCq=q^{3}+3q^{2}+12q+300$$

The equation is in standard form.

$$3CqT=q^{3}+3q^{2}+12q+300$$

Divide both sides by 3Cq.

$$\frac{3CqT}{3Cq}=\frac{q^{3}+3q^{2}+12q+300}{3Cq}$$

Dividing by 3Cq undoes the multiplication by 3Cq.

$$T=\frac{q^{3}+3q^{2}+12q+300}{3Cq}$$