To solve the system of linear equations by substitution and determine whether the system has one solution, no solution, or an infinite number of solutions, follow these steps:

Given the system of equations:

1. $x + y = 6$
2. $-x + y = -2$

First, solve Equation 1 for $y$:

$y = 6 - x \quad \text{(Equation 3)}$

Next, substitute Equation 3 into the second equation:

$-x + y = -2$

Substitute $y = 6 - x$:

$-x + (6 - x) = -2$

Simplify and solve for $x$:

$-x + 6 - x = -2$

$-2x + 6 = -2$

Subtract 6 from both sides:

$-2x = -8$

Divide by -2:

$x = 4$

Now, substitute $x = 4$ back into Equation 3 to find $y$:

$y = 6 - 4$

$y = 2$

Therefore, the solution to the system of equations is:

$(x, y) = (4, 2)$

To verify, substitute $x = 4$ and $y = 2$ back into the original equations:

1. $x + y = 6$

$4 + 2 = 6$

This is true.

2. $-x + y = -2$

$-4 + 2 = -2$

This is also true.

Thus, the system has one solution: $(x, y) = (4, 2)$.

Do you want more details or have any questions?

Here are some related questions:

1. What are the steps for solving a system of equations by elimination?
2. How can you determine if a system of linear equations has no solution?
3. How can you determine if a system of linear equations has infinitely many solutions?
4. What are some common methods for solving systems of linear equations?
5. How do you graphically represent the solution of a system of linear equations?
6. What is the importance of substitution in solving linear systems?
7. How can matrices be used to solve systems of linear equations?
8. What are real-life applications of solving systems of linear equations?

Tip: Always check your solution by substituting the values back into the original equations to verify accuracy.