Question

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

Short answer

x = -41/7, y = 10/7

Step-by-step solution

Separate the real and imaginary parts

Given the equation ewline (x + 2y + 3) + i(3x - y - 1) = 0 ewline The sum of real and imaginary parts of the equation is equal to zero. Therefore, we set the real part equal to zero and the imaginary part equal to zero.

Set the real part to zero

The real part of the equation is: (x + 2y + 3) = 0 ewline Solve for x: x + 2y + 3 = 0

Set the imaginary part to zero

The imaginary part of the equation is: (3x - y - 1) = 0 ewline Solve for y: 3x - y - 1 = 0

Solve the system of linear equations

We now have two linear equations: ewline 1) x + 2y + 3 = 0 2) 3x - y - 1 = 0 Use substitution or elimination method to solve for x and y.

Solve the first equation for x or y

Solve the first equation for x: ewlinex = -2y - 3

Substitute into the second equation

Substitute x = -2y - 3 into the second equation: 3(-2y - 3) - y - 1 = 0 Simplify: -6y - 9 - y - 1 = 0 -7y - 10 = 0

Solve for y

Solve for y: -7y - 10 = 0 ewline y = -10 /-7 ewline y = 10/7

Solve for x

Substitute y = 10/7 back into x = -2y - 3: x = -2(10/7) - 3 = -20/7 - 21/7 = -41/7

Key concepts

Real and Imaginary Parts

In complex equations, numbers are often expressed as a combination of real and imaginary parts. The general form is:
a + bi = 0,
where 'a' is the real part, 'b' is the coefficient of the imaginary part 'i', and 'i' is the imaginary unit satisfying i² = -1.
To solve such equations, we separate the equation into two parts: the real part and the imaginary part. We set each part equal to zero since the sum of a real and imaginary number can only be zero if both parts are zero independently. In our problem, this results in:

• Real part: x + 2y + 3 = 0
• Imaginary part: 3x - y - 1 = 0

System of Linear Equations

A system of linear equations is a set of equations with multiple variables that can be solved together. In our problem, separating the real and imaginary parts gave us two linear equations:
1. x + 2y + 3 = 0 2. 3x - y - 1 = 0
These equations must both be satisfied by the values of x and y. The system can be written in matrix form or solved using methods such as substitution or elimination. The goal is to find the values of the variables that make both equations true.
Understanding this concept is crucial, as it applies across a wide range of mathematical problems beyond complex numbers.

Substitution Method

The substitution method involves solving one of the equations for one variable in terms of the other variable, and then substituting this expression into the other equation. It's a step-by-step process:
1. Solve one of the equations for one variable, for instance, solving equation (1) for x: x = -2y - 3.
2. Substitute this expression into the other equation to replace x. This turns the second equation into an equation with just one variable (y): 3(-2y - 3) - y - 1 = 0.
3. Solve this new equation for the remaining variable. This method simplifies the process by reducing the number of variables in the equations, making it easier to find a solution.

Elimination Method

The elimination method, also known as the addition method, involves adding or subtracting equations to eliminate one of the variables. Here's how it works:
1. Align the equations, ensuring like terms are in columns. For instance, we have:

• x + 2y + 3 = 0
• 3x - y - 1 = 0

2. Multiply one or both equations by constants if necessary, so the coefficients of one of the variables are opposites. 3. Add or subtract the equations to cancel one variable. For instance:If we multiply the first equation by 3, we get: 3x + 6y + 9 = 0,
and the second equation remains: 3x - y - 1 = 0.
Subtracting these equations results in: 7y + 10 = 0, which can easily be solved for y.
4. Substitute the value obtained back into one of the original equations to solve for the other variable. This method is particularly useful when the system of equations aligns well for elimination.