Solve (2-3i)(x+yi)=4+i

$$( 2 - 3 i ) ( x + y i ) = 4 + i$$

Solve for x

$$x=\frac{5}{13}+\frac{14}{13}i-iy$$

Solution Steps

Divide both sides by 2-3i.

$$x+yi=\frac{4+i}{2-3i}$$

Multiply both numerator and denominator of \frac{4+i}{2-3i} by the complex conjugate of the denominator, 2+3i.

$$x+yi=\frac{\left(4+i\right)\left(2+3i\right)}{\left(2-3i\right)\left(2+3i\right)}$$

Do the multiplications in \frac{\left(4+i\right)\left(2+3i\right)}{\left(2-3i\right)\left(2+3i\right)}.

$$x+yi=\frac{5+14i}{13}$$

Divide 5+14i by 13 to get \frac{5}{13}+\frac{14}{13}i.

$$x+yi=\frac{5}{13}+\frac{14}{13}i$$

Subtract yi from both sides.

$$x=\frac{5}{13}+\frac{14}{13}i-yi$$

Multiply -1 and i to get -i.

$$x=\frac{5}{13}+\frac{14}{13}i-iy$$

Solve for y

$$y=ix+\left(\frac{14}{13}-\frac{5}{13}i\right)$$

Steps for Solving Linear Equation

Divide both sides by 2-3i.

$$x+yi=\frac{4+i}{2-3i}$$

Multiply both numerator and denominator of \frac{4+i}{2-3i} by the complex conjugate of the denominator, 2+3i.

$$x+yi=\frac{\left(4+i\right)\left(2+3i\right)}{\left(2-3i\right)\left(2+3i\right)}$$

Do the multiplications in \frac{\left(4+i\right)\left(2+3i\right)}{\left(2-3i\right)\left(2+3i\right)}.

$$x+yi=\frac{5+14i}{13}$$

Divide 5+14i by 13 to get \frac{5}{13}+\frac{14}{13}i.

$$x+yi=\frac{5}{13}+\frac{14}{13}i$$

Subtract x from both sides.

$$yi=\frac{5}{13}+\frac{14}{13}i-x$$

The equation is in standard form.

$$iy=\frac{5}{13}+\frac{14}{13}i-x$$

Divide both sides by i.

$$\frac{iy}{i}=\frac{\frac{5}{13}+\frac{14}{13}i-x}{i}$$

Dividing by i undoes the multiplication by i.

$$y=\frac{\frac{5}{13}+\frac{14}{13}i-x}{i}$$

Divide \frac{5}{13}+\frac{14}{13}i-x by i.

$$y=ix+\left(\frac{14}{13}-\frac{5}{13}i\right)$$