To solve the given system of equations:
- \( 3x + 2y = 15 \) (Equation 1)
- \( 5x - 4y = -8 \) (Equation 2)
Let's use the elimination method to solve for \(x\) and \(y\).
First, we can multiply Equation 1 by 2 in order to align the \(y\) coefficients:
\[ 2(3x + 2y) = 2(15) \implies 6x + 4y = 30 \quad (Equation 3) \]
Now we have:
- Equation 3: \( 6x + 4y = 30 \)
- Equation 2: \( 5x - 4y = -8 \)
Next, we can add Equation 3 and Equation 2:
\[ (6x + 4y) + (5x - 4y) = 30 + (-8) \] \[ 6x + 4y + 5x - 4y = 30 - 8 \] \[ 11x = 22 \]
Now we solve for \(x\):
\[ x = \frac{22}{11} = 2 \]
Next, we substitute \(x = 2\) back into Equation 1 to find \(y\):
\[ 3(2) + 2y = 15 \] \[ 6 + 2y = 15 \] \[ 2y = 15 - 6 \] \[ 2y = 9 \] \[ y = \frac{9}{2} \]
The solution to the system of equations is \( (2, \frac{9}{2}) \).
Thus, the answers to the questions are:
- The elimination method was used to solve the system. (Response A)
- The solution to the system is \( (2, \frac{9}{2}) \). (Response B)
2 answers