Solve 3+xi/1+2i=y+2i

$$\frac { 3 + x i } { 1 + 2 i } = y + 2 i$$

Solve for ξ

$$\xi =\left(1+2i\right)y+\left(-7+2i\right)$$

Steps for Solving Linear Equation

Divide each term of 3+\xi by 1+2i to get \frac{3}{1+2i}+\frac{\xi }{1+2i}.

$$\frac{3}{1+2i}+\frac{\xi }{1+2i}=y+2i$$

Multiply both numerator and denominator of \frac{3}{1+2i} by the complex conjugate of the denominator, 1-2i.

$$\frac{3\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}+\frac{\xi }{1+2i}=y+2i$$

Do the multiplications in \frac{3\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}.

$$\frac{3-6i}{5}+\frac{\xi }{1+2i}=y+2i$$

Divide 3-6i by 5 to get \frac{3}{5}-\frac{6}{5}i.

$$\frac{3}{5}-\frac{6}{5}i+\frac{\xi }{1+2i}=y+2i$$

Subtract \frac{3}{5}-\frac{6}{5}i from both sides.

$$\frac{\xi }{1+2i}=y+2i-\left(\frac{3}{5}-\frac{6}{5}i\right)$$

Multiply -1 and \frac{3}{5}-\frac{6}{5}i to get -\frac{3}{5}+\frac{6}{5}i.

$$\frac{\xi }{1+2i}=y+2i+\left(-\frac{3}{5}+\frac{6}{5}i\right)$$

Do the additions in 2i+\left(-\frac{3}{5}+\frac{6}{5}i\right).

$$\frac{\xi }{1+2i}=y-\frac{3}{5}+\frac{16}{5}i$$

The equation is in standard form.

$$\left(\frac{1}{5}-\frac{2}{5}i\right)\xi =y+\left(-\frac{3}{5}+\frac{16}{5}i\right)$$

Divide both sides by \frac{1}{5}-\frac{2}{5}i.

$$\frac{\left(\frac{1}{5}-\frac{2}{5}i\right)\xi }{\frac{1}{5}-\frac{2}{5}i}=\frac{y+\left(-\frac{3}{5}+\frac{16}{5}i\right)}{\frac{1}{5}-\frac{2}{5}i}$$

Dividing by \frac{1}{5}-\frac{2}{5}i undoes the multiplication by \frac{1}{5}-\frac{2}{5}i.

$$\xi =\frac{y+\left(-\frac{3}{5}+\frac{16}{5}i\right)}{\frac{1}{5}-\frac{2}{5}i}$$

Divide y+\left(-\frac{3}{5}+\frac{16}{5}i\right) by \frac{1}{5}-\frac{2}{5}i.

$$\xi =\left(1+2i\right)y+\left(-7+2i\right)$$

Solve for y

$$y=\left(\frac{1}{5}-\frac{2}{5}i\right)\xi +\left(\frac{3}{5}-\frac{16}{5}i\right)$$