The problem asks us to determine in which quadrant the solution of the given system of linear equations lies. The equations are:

1. $x - 3y = 6$
2. $x + y = 2$

Step 1: Solving the System of Equations

We can solve the system using the substitution or elimination method. Let’s solve using substitution:

1. From the second equation:
   $x + y = 2 \implies y = 2 - x$

2. Substitute $y = 2 - x$ into the first equation:
   $x - 3(2 - x) = 6$

3. Simplify the equation:
   $x - 6 + 3x = 6 \implies 4x - 6 = 6$

4. Solving for $x$:
   $4x = 12 \implies x = 3$

5. Substitute $x = 3$ back into $y = 2 - x$:
   $y = 2 - 3 = -1$

Step 2: Determining the Quadrant

The solution is $(3, -1)$. In the coordinate plane:

• $x = 3$ is positive.
• $y = -1$ is negative.

This corresponds to Quadrant IV, where $x > 0$ and $y < 0$.

Final Answer

The solution to the system of equations lies in Quadrant IV.

Would you like a detailed explanation of any step or have any follow-up questions?

Related Questions

1. What method can be used to solve a system of linear equations besides substitution?
2. How can you determine the quadrant of a point without plotting it?
3. What happens when a system of linear equations has no solution?
4. How do you graphically find the intersection of two linear equations?
5. What do parallel lines in a system of equations signify?

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