Let's solve the given system of equations step by step as per your instructions.
### Given Equations:
1. \( 3x + 4y = 48 \) (Equation 1)
2. \( 6x + 2y = 60 \) (Equation 2)
### 1. Method of Choice:
**Choose a method:**
**elimination**
This method is suitable since the equations are set up so that we can easily manipulate them to cancel out one variable.
### 2. Getting the x variables to cancel out:
**Operation to get the x variables to cancel:**
Multiply the top equation by -2 so that the \( x \) terms in both equations will cancel out when added:
\[
-2(3x + 4y = 48) \Rightarrow -6x - 8y = -96
\]
Now, we can use this to cancel with the second equation.
### 3. Simplified top equation:
The simplified top equation from the operation is:
**\(-6x - 8y = -96\)**
### 4. Line them up and add the equations together:
Now we need to line up the modified top equation with the bottom equation:
\[
\begin{align*}
-6x - 8y & = -96 \quad \text{(Modified Equation 1)} \\
6x + 2y & = 60 \quad \text{(Equation 2)} \\
\end{align*}
\]
Adding these two equations together:
\[
(-6x + 6x) + (-8y + 2y) = -96 + 60
\]
This simplifies to:
\[
0x - 6y = -36
\]
### 5. Solve for y:
To solve for \( y \):
\[
-6y = -36 \\
y = \frac{-36}{-6} = 6
\]
### 6. Solve for x:
Now we can substitute \( y = 6 \) back into one of the original equations, let's use Equation 1:
\[
3x + 4(6) = 48 \\
3x + 24 = 48 \\
3x = 48 - 24 \\
3x = 24 \\
x = \frac{24}{3} = 8
\]
### Final Solution:
\( x = 8 \) and \( y = 6 \).
Thus, the solution to the system of equations is \( (x, y) = (8, 6) \).