How To Solve The Systems Of Equations: 1. { X + 2y + 1 = 0, 2x - 3y - 12 = 0 } 2. { 3x - 2y + 3 = 0, 4x + 3y - 47 = 0 } 3. { 3x + 2y + 25 = 0, 2x + Y + 10 = 0 }?

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This article delves into the methods for solving systems of linear equations, a fundamental concept in algebra. We'll explore various techniques for finding solutions, focusing on three specific examples. Understanding these methods is crucial for success in various mathematical and scientific disciplines. This guide provides step-by-step solutions and explanations, ensuring a clear understanding of the underlying principles.

Methods for Solving Systems of Equations

Before we dive into specific examples, let's briefly discuss the common methods for solving systems of equations:

  • Substitution: This method involves solving one equation for one variable and substituting that expression into the other equation. This reduces the system to a single equation with one variable, which can then be easily solved. Once one variable is found, its value is substituted back into either original equation to find the other variable.
  • Elimination (or Addition/Subtraction): This method involves manipulating the equations (by multiplying them by constants) so that the coefficients of one variable are opposites. Adding the equations then eliminates that variable, leaving a single equation with one variable. Solving this equation and substituting back yields the solution.
  • Graphing: This method involves plotting the equations on a coordinate plane. The point(s) of intersection represent the solution(s) to the system. While visually intuitive, this method is most accurate for systems with integer solutions and can be less precise for non-integer solutions.

We will primarily use the substitution and elimination methods in the following examples, as they are generally more efficient and accurate for algebraic solutions.

Example 1: Solving a System Using Substitution

Let's tackle the first system of equations:

\begin{cases}
x + 2y + 1 = 0 \\
2x - 3y - 12 = 0
\end{cases}

Our primary goal in solving this system is to find the values of x and y that satisfy both equations simultaneously. We will employ the substitution method here. The key is to isolate one variable in one equation and then substitute that expression into the other equation. This will leave us with a single equation with a single variable, which we can then easily solve.

First, we'll rearrange the first equation to solve for x:

x + 2y + 1 = 0
x = -2y - 1

Now that we have x expressed in terms of y, we will substitute this expression into the second equation:

2x - 3y - 12 = 0
2(-2y - 1) - 3y - 12 = 0

Distribute the 2:

-4y - 2 - 3y - 12 = 0

Combine like terms:

-7y - 14 = 0

Add 14 to both sides:

-7y = 14

Divide both sides by -7:

y = -2

We have now found the value of y. To find the value of x, we'll substitute y = -2 back into the equation x = -2y - 1:

x = -2(-2) - 1
x = 4 - 1
x = 3

Therefore, the solution to the system of equations is x = 3 and y = -2. We can verify this solution by plugging these values back into the original equations:

  • Equation 1: 3 + 2(-2) + 1 = 3 - 4 + 1 = 0 (Correct)
  • Equation 2: 2(3) - 3(-2) - 12 = 6 + 6 - 12 = 0 (Correct)

Thus, our solution x = 3 and y = -2 is correct. This substitution process exemplifies a fundamental algebraic technique for solving systems of equations.

Example 2: Solving a System Using Elimination

Let's consider the second system of equations:

\begin{cases}
3x - 2y + 3 = 0 \\
4x + 3y - 47 = 0
\end{cases}

In this case, we will use the elimination method to solve the system. The elimination method, also sometimes referred to as the addition or subtraction method, is particularly useful when the equations are structured in a way that allows us to cancel out one of the variables. The core strategy behind the elimination method is to manipulate the equations so that the coefficients of one of the variables are opposites (i.e., one is the negative of the other). When the equations are added together, this variable will be eliminated, leaving us with a single equation in a single variable.

To eliminate y, we need to make the coefficients of y in both equations opposites. We can achieve this by multiplying the first equation by 3 and the second equation by 2:

3(3x - 2y + 3) = 0  =>  9x - 6y + 9 = 0
2(4x + 3y - 47) = 0 => 8x + 6y - 94 = 0

Now, the coefficients of y are -6 and +6, which are opposites. Adding the two modified equations eliminates y:

(9x - 6y + 9) + (8x + 6y - 94) = 0
17x - 85 = 0

Add 85 to both sides:

17x = 85

Divide both sides by 17:

x = 5

Now that we have found the value of x, we substitute x = 5 back into either of the original equations to solve for y. Let's use the first equation:

3x - 2y + 3 = 0
3(5) - 2y + 3 = 0
15 - 2y + 3 = 0
18 - 2y = 0

Subtract 18 from both sides:

-2y = -18

Divide both sides by -2:

y = 9

Therefore, the solution to the system of equations is x = 5 and y = 9. We can verify this solution by plugging these values back into the original equations:

  • Equation 1: 3(5) - 2(9) + 3 = 15 - 18 + 3 = 0 (Correct)
  • Equation 2: 4(5) + 3(9) - 47 = 20 + 27 - 47 = 0 (Correct)

Thus, the elimination method has efficiently provided us with the correct solution x = 5 and y = 9, further demonstrating the technique's utility in solving systems of equations.

Example 3: Solving a System with Rearrangement and Substitution

Finally, let's consider the third system of equations:

\begin{cases}
3x + 2y + 25 = 0 \\
2x + y + 10 = 0
\end{cases}

For this system, we will again employ the substitution method, but we will also need to rearrange one of the equations to make the substitution process easier. Strategic rearrangement is a vital part of solving systems effectively. The goal is to isolate one variable in one of the equations, ideally a variable with a coefficient of 1 or -1 to simplify calculations.

Looking at the second equation, we can easily isolate y:

2x + y + 10 = 0
y = -2x - 10

Now, we substitute this expression for y into the first equation:

3x + 2y + 25 = 0
3x + 2(-2x - 10) + 25 = 0

Distribute the 2:

3x - 4x - 20 + 25 = 0

Combine like terms:

-x + 5 = 0

Subtract 5 from both sides:

-x = -5

Multiply both sides by -1:

x = 5

Now that we have found x = 5, we substitute this value back into the equation y = -2x - 10:

y = -2(5) - 10
y = -10 - 10
y = -20

Therefore, the solution to this system of equations is x = 5 and y = -20. Let's verify this solution by substituting these values into the original equations:

  • Equation 1: 3(5) + 2(-20) + 25 = 15 - 40 + 25 = 0 (Correct)
  • Equation 2: 2(5) + (-20) + 10 = 10 - 20 + 10 = 0 (Correct)

Our solution x = 5 and y = -20 is indeed correct. This example highlights the importance of strategic equation rearrangement in conjunction with the substitution method for efficiently solving systems of equations.

Conclusion

This article has demonstrated the process of solving systems of equations using both the substitution and elimination methods. Understanding these techniques is crucial for various applications in mathematics, science, and engineering. By mastering these methods, you can effectively solve a wide range of problems involving multiple variables and constraints. Remember to practice consistently to develop your skills and confidence in solving systems of equations.