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Linear algebra is a branch of mathematics that deals with vectors, vector spaces, linear equations, and matrices. One of the fundamental tools in linear algebra is the system of linear equations—a collection of linear equations involving the same set of variables. Solving these systems can help us find the values of the variables that satisfy all the given equations. Understanding the core concepts of linear algebra is essential for solving real-world problems efficiently. Linear algebra has applications in various fields like computer science, engineering, physics, and economics. In this exercise, we are dealing with a system of linear equations, and our goal is to find the values of x and y that satisfy all the given equations simultaneously.

Solving equations is an essential skill in mathematics, particularly in algebra. It involves finding the values of variables that satisfy given mathematical statements. There are multiple methods to solve equations, including graphing, elimination, and substitution. In the case of a system of linear equations, we aim to find a common solution for all the equations in the system. This step-by-step process typically involves:

• Labeling the given equations for clarity.
• Simplifying equations if possible.
• Isolating one variable to make substitution easier.
• Solving for one variable and then using that value to find the other variables.
• Finally, verifying the solution by substituting the values back into the original equations.

These steps not only help us find the solution but also ensure that the solution is correct.

The substitution method is a common technique for solving systems of linear equations. This method involves solving one equation for one variable and then substituting that expression into the other equation(s). Here is a detailed breakdown of the method:

1. **Solve one equation for one of the variables.** For example, in the exercise, we solve Equation (3) for x: \(x = 11 - y\).

2. **Substitute the expression into the other equation.** Using the expression from the first step, substitute into another equation to solve for the second variable. From the exercise, substituting \(x = 11 - y\) into Equation (4) yields \(-(11 - y) + 2y = 7\).

3. **Solve for the second variable.** After substitution, simplify the equation to find the value of the second variable. Here, solving \(-11 + 3y = 7\) results in \(y = 6\).

4. **Use the known value to find the other variable.** Substitute the known value back into the expression for the first variable. In the exercise, substituting \(y = 6\) into \(x = 11 - y\) gives \(x = 5\).

The substitution method is particularly useful when one of the equations is easily solvable for one variable. It simplifies the process of finding the solution to the system.

Verification of solutions is an important final step when solving systems of linear equations. It ensures that the obtained solutions satisfy all the original equations. To verify the solutions, substitute the values of the variables back into each equation and check if both sides are equal. For the exercise we solved:

1. **Substitute into the first equation.** For Equation (1): \(3x - 2y = 3\), substitute \(x = 5\) and \(y = 6\). This gives \(3(5) - 2(6) = 3\) which simplifies to \(15 - 12 = 3\)—true.

2. **Substitute into the simplified second equation.** For Equation (4), \(-x + 2y = 7\), substitute \(x = 5\) and \(y = 6\). This gives \(-5 + 2(6) = 7\) which simplifies to \(-5 + 12 = 7\)—true.

By verifying both equations, we can confidently conclude that our solutions \(x = 5\) and \(y = 6\) are correct. This step is crucial for ensuring the accuracy and reliability of our results in mathematical problem-solving.