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To solve this system of linear equations:

1. \( y = -\frac{1}{3}x + 2 \)
2. \( y = -\frac{2}{3}x + 4 \)

We can solve this by setting the right-hand sides of the two equations equal to each other, since both equal \( y \):

\[
-\frac{1}{3}x + 2 = -\frac{2}{3}x + 4
\]

### Step 1: Eliminate fractions
Multiply through by 3 to clear the fractions:

\[
-1x + 6 = -2x + 12
\]

### Step 2: Simplify the equation
Add \( 2x \) to both sides:

\[
x + 6 = 12
\]

### Step 3: Solve for \( x \)
Subtract 6 from both sides:

\[
x = 6
\]

### Step 4: Substitute \( x = 6 \) back into one of the original equations
We'll use the first equation \( y = -\frac{1}{3}x + 2 \):

\[
y = -\frac{1}{3}(6) + 2
\]
\[
y = -2 + 2 = 0
\]

### Solution:
The solution to the system is \( (6, 0) \).

Would you like further details or have any questions?

Here are some related questions:
1. How do you solve a system of equations using substitution?
2. What is the graphical representation of the solution to this system?
3. How do you solve a system of equations when the lines are parallel?
4. What happens if the system has no solution?
5. How can you check the solution for correctness?

**Tip:** Always try to eliminate fractions early on by multiplying through by a common denominator.

Solve the system of equations: y = -1/3x + 2 and y = -2/3x + 4.

Learn how to solve the system of equations y = -1/3x + 2 and y = -2/3x + 4 using the substitution method. This problem provides a clear solution step-by-step for students in Grades 7-9, demonstrating key algebraic concepts like solving systems of linear equations.

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