To solve the system of equations
- \( x - 3y = -10 \) (Equation 1)
- \( 3x + 2y = 14 \) (Equation 2)
we can use the substitution or elimination method. Here, we'll use the substitution method.
First, we'll solve Equation 1 for \( x \):
\[ x = 3y - 10 \]
Next, we'll substitute this expression for \( x \) into Equation 2:
\[ 3(3y - 10) + 2y = 14 \]
Now, distribute the 3:
\[ 9y - 30 + 2y = 14 \]
Combine like terms:
\[ 11y - 30 = 14 \]
Next, add 30 to both sides:
\[ 11y = 44 \]
Now, divide by 11:
\[ y = 4 \]
Now that we have the value of \( y \), we can substitute it back into the expression we found for \( x \):
\[ x = 3(4) - 10 \]
\[ x = 12 - 10 \]
\[ x = 2 \]
So the solution to the system of equations is:
\[ x = 2, \quad y = 4 \]
We can write the solution as the ordered pair:
\[ (2, 4) \]
To confirm, we can substitute \( x = 2 \) and \( y = 4 \) back into the original equations:
-
For Equation 1: \[ 2 - 3(4) = 2 - 12 = -10 \quad \text{(True)} \]
-
For Equation 2: \[ 3(2) + 2(4) = 6 + 8 = 14 \quad \text{(True)} \]
Both equations are satisfied. Thus, the solution is confirmed:
\[ \boxed{(2, 4)} \]