Solve (a^2-b)i+2abj=3i+4j

$$( a ^ { 2 } - b ) i + 2 a b j = 3 i + 4 j$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{ia^{2}-4j-3i}{2aj-i}\text{, }&j=0\text{ or }a\neq \frac{i}{2j}\\b\in \mathrm{C}\text{, }&\left(j=\frac{1}{4}i\text{ and }a=2\right)\text{ or }\left(j=-\frac{1}{2}i\text{ and }a=-\frac{2\times 3}{3+3\sqrt{3}i}-\frac{\sqrt{3}i}{2}-\frac{1}{2}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a^{2}-b by i.

$$ia^{2}-ib+2abj=3i+4j$$

Subtract ia^{2} from both sides.

$$-ib+2abj=3i+4j-ia^{2}$$

Reorder the terms.

$$2abj-ib=-ia^{2}+4j+3i$$

Combine all terms containing b.

$$\left(2aj-i\right)b=-ia^{2}+4j+3i$$

The equation is in standard form.

$$\left(2aj-i\right)b=3i+4j-ia^{2}$$

Divide both sides by -i+2aj.

$$\frac{\left(2aj-i\right)b}{2aj-i}=\frac{3i+4j-ia^{2}}{2aj-i}$$

Dividing by -i+2aj undoes the multiplication by -i+2aj.

$$b=\frac{3i+4j-ia^{2}}{2aj-i}$$

Solve for a

$$a=-i\sqrt{-b+4ij+\left(bj\right)^{2}-3}+ibj$$

$$a=i\left(\sqrt{-b+4ij+\left(bj\right)^{2}-3}+bj\right)$$