Question

Solve for all possible values of the real numbers $\mathit{x}$and $\mathit{y}$in the following equations.$\left(x+2y+3\right)\mathbf{+}\mathit{i}\left(3x-y-1\right)\mathbf{=}\mathbf{0}$

Short answer

The answer is obtained:

$x=-\frac{1}{7},y=-\frac{10}{7}$

Step-by-step solution

Given information

It is given that $(\mathrm{x}+2\mathrm{y}+3)+i(3\mathrm{x}-\mathrm{y}-1)=0$.

Definition of a complex number

Every complex number can be expressed as:

$z=a+bi$

Whereaandbare both real numbers, z is the complex number, and i is known as iota, which makes z a complex number.

Begin by stating the given information

State the given information:

$(\mathrm{x}+2\mathrm{y}+3)+i(3\mathrm{x}-\mathrm{y}-1)=0$

Equate like terms

Equate like terms from both sides as:

$$\begin{gathered} x+2y=-3 \\ 3x-y=1 \end{gathered}$$

Solve the two equations to find the solutions

Multiply the second equation by$2$and add it to the first equation as:

$$\begin{gathered} x+2y=-3 \\ 6x-2y=2 \end{gathered}$$

Continue evaluation:

$$\begin{gathered} 7x=-1 \\ x=-\frac{1}{7} \\ \mathrm{and} \\ \mathrm{y}=3\mathrm{x}-1 \\ \mathrm{y}=3\left(-\frac{1}{7}\right)-1 \\ \mathrm{y}=-\frac{3}{7}-1 \\ \mathrm{y}=-\frac{10}{7} \end{gathered}$$

Thus, the final answer is obtained:

$x=-\frac{1}{7},y=-\frac{10}{7}$