Q41P - Mathematical Methods in Physical Sciences

Question

Solve for all possible values of the real numbers $\mathit{x}$ and $\mathit{y}$ in the following equations. $\left(2x-3y-5\right)\mathbf{+}\mathit{i}\left(x+2y+1\right)\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The answer is obtained:

 $x=1,y=-1$

Step 1: Given information

It is given that $\left(2x-3y-5\right)+i\left(x+2y+1\right)=0$.

Step 2: Definition of a complex number

Every complex number can be expressed as: 

 

z=a+bi

 

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

Step 3: Begin by stating the given information

State the given information:

 

$\left(2x-3y-5\right)+i\left(x+2y+1\right)=0$

Step 4: Equate like terms

Equate like terms from both sides:

 $$\begin{gathered} 2x-3y=5 \\ x+2y=-1 \end{gathered}$$

Step 5: Solve the two equations to find the solutions

Multiply the second equation by $2$ and subtract it from the first equation:

 

$$\begin{gathered} 2x-3y=5 \\ 2x+4y=-2 \end{gathered}$$

 

Continue evaluation as:

 

$$\begin{gathered} -7y=7 \\ y=-1 \\ \mathrm{and} \\ 2\mathrm{x}=5+3\left(-1\right) \\ \mathrm{x}=1 \end{gathered}$$

 

Thus, the final answer is obtained:

 $x=1,y=-1$