Sure! We can substitute \( y = 3x - 2 \) into the second equation \( x = -2y + 10 \).
Here’s how it looks:
Starting with the second equation:
\[ x = -2y + 10 \]
Substituting \( y = 3x - 2 \) into this equation gives:
\[ x = -2(3x - 2) + 10 \]
Now, let's simplify that:
\[ x = -6x + 4 + 10 \]
Combine like terms:
\[ x = -6x + 14 \]
Now, add \( 6x \) to both sides:
\[ x + 6x = 14 \]
This simplifies to:
\[ 7x = 14 \]
Now, divide both sides by 7:
\[ x = 2 \]
Now that we have \( x \), we can substitute back to find \( y \):
Using the first equation \( y = 3x - 2 \):
\[ y = 3(2) - 2 = 6 - 2 = 4 \]
So the solution to the system of equations is:
\[ x = 2, \quad y = 4 \]