Solve 4(BL^2+CM^2)=5BC^2

$$4 ( BL ^ { 2 } + CM ^ { 2 } ) = 5 BC ^ { 2 }$$

Solve for B

$$\left\{\begin{matrix}B=-\frac{4CM^{2}}{4L^{2}-5C^{2}}\text{, }&|L|\neq \frac{\sqrt{5}|C|}{2}\\B\in \mathrm{R}\text{, }&\left(C=0\text{ and }L=0\right)\text{ or }\left(M=0\text{ and }|L|=\frac{\sqrt{5}|C|}{2}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 4 by BL^{2}+CM^{2}.

$$4BL^{2}+4CM^{2}=5BC^{2}$$

Subtract 5BC^{2} from both sides.

$$4BL^{2}+4CM^{2}-5BC^{2}=0$$

Subtract 4CM^{2} from both sides. Anything subtracted from zero gives its negation.

$$4BL^{2}-5BC^{2}=-4CM^{2}$$

Combine all terms containing B.

$$\left(4L^{2}-5C^{2}\right)B=-4CM^{2}$$

Divide both sides by -5C^{2}+4L^{2}.

$$\frac{\left(4L^{2}-5C^{2}\right)B}{4L^{2}-5C^{2}}=-\frac{4CM^{2}}{4L^{2}-5C^{2}}$$

Dividing by -5C^{2}+4L^{2} undoes the multiplication by -5C^{2}+4L^{2}.

$$B=-\frac{4CM^{2}}{4L^{2}-5C^{2}}$$

Solve for C

$$\left\{\begin{matrix}C=-\frac{2\left(\sqrt{M^{4}+5\left(BL\right)^{2}}-M^{2}\right)}{5B}\text{; }C=\frac{2\left(\sqrt{M^{4}+5\left(BL\right)^{2}}+M^{2}\right)}{5B}\text{, }&B\neq 0\\C=0\text{, }&B=0\text{ and }M\neq 0\\C\in \mathrm{R}\text{, }&B=0\text{ and }M=0\end{matrix}\right.$$