Solve z+3(1+i)z-8(2-i)=0

$$z + 3 ( 1 + i ) z - 8 ( 2 - i ) = 0$$

Solve for z

$$z=\frac{8}{5}-\frac{16}{5}i=1.6-3.2i$$

Steps for Solving Linear Equation

Multiply 3 times 1+i.

$$z+\left(3\times 1+3i\right)z-8\left(2-i\right)=0$$

Do the multiplications in 3\times 1+3i.

$$z+\left(3+3i\right)z-8\left(2-i\right)=0$$

Combine z and \left(3+3i\right)z to get \left(4+3i\right)z.

$$\left(4+3i\right)z-8\left(2-i\right)=0$$

Multiply 8 times 2-i.

$$\left(4+3i\right)z-\left(8\times 2+8\left(-i\right)\right)=0$$

Do the multiplications in 8\times 2+8\left(-i\right).

$$\left(4+3i\right)z-\left(16-8i\right)=0$$

Add 16-8i to both sides.

$$\left(4+3i\right)z=0+\left(16-8i\right)$$

Anything plus zero gives itself.

$$\left(4+3i\right)z=16-8i$$

Divide both sides by 4+3i.

$$z=\frac{16-8i}{4+3i}$$

Multiply both numerator and denominator of \frac{16-8i}{4+3i} by the complex conjugate of the denominator, 4-3i.

$$z=\frac{\left(16-8i\right)\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$z=\frac{\left(16-8i\right)\left(4-3i\right)}{4^{2}-3^{2}i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$z=\frac{\left(16-8i\right)\left(4-3i\right)}{25}$$

Multiply complex numbers 16-8i and 4-3i like you multiply binomials.

$$z=\frac{16\times 4+16\times \left(-3i\right)-8i\times 4-8\left(-3\right)i^{2}}{25}$$

By definition, i^{2} is -1.

$$z=\frac{16\times 4+16\times \left(-3i\right)-8i\times 4-8\left(-3\right)\left(-1\right)}{25}$$

Do the multiplications in 16\times 4+16\times \left(-3i\right)-8i\times 4-8\left(-3\right)\left(-1\right).

$$z=\frac{64-48i-32i-24}{25}$$

Combine the real and imaginary parts in 64-48i-32i-24.

$$z=\frac{64-24+\left(-48-32\right)i}{25}$$

Do the additions in 64-24+\left(-48-32\right)i.

$$z=\frac{40-80i}{25}$$

Divide 40-80i by 25 to get \frac{8}{5}-\frac{16}{5}i.

$$z=\frac{8}{5}-\frac{16}{5}i$$