To solve the system of equations given by \(3x + 4y = 48\) and \(6x + 2y = 60\), we can use the method of substitution or elimination. Here, we will use the elimination method for clarity.
Step 1: Write down the equations
- \(3x + 4y = 48\) , (Equation 1)
- \(6x + 2y = 60\) , (Equation 2)
Step 2: Simplify Equation 2
Notice that Equation 2 can be simplified by dividing all terms by 2:
\[ 3x + y = 30 \]
Now we have a new system:
- \(3x + 4y = 48\) , (Equation 1)
- \(3x + y = 30\) , (Equation 3)
Step 3: Eliminate \(x\)
We can eliminate \(x\) by subtracting Equation 3 from Equation 1.
\[ (3x + 4y) - (3x + y) = 48 - 30 \]
This simplifies to:
\[ 4y - y = 18 \implies 3y = 18 \]
Step 4: Solve for \(y\)
Divide both sides by 3:
\[ y = 6 \]
Step 5: Substitute \(y\) back into one of the equations
We can substitute \(y = 6\) back into Equation 3 to solve for \(x\):
\[ 3x + 6 = 30 \]
Subtract 6 from both sides:
\[ 3x = 24 \]
Divide by 3:
\[ x = 8 \]
Step 6: Write the solution
The solution to the system of equations is:
\[ (x, y) = (8, 6) \]
Step 7: Check the solution
To ensure our solution is correct, we can substitute \(x = 8\) and \(y = 6\) back into the original equations.
For Equation 1:
\[ 3(8) + 4(6) = 24 + 24 = 48 \quad \text{(True)} \]
For Equation 2:
\[ 6(8) + 2(6) = 48 + 12 = 60 \quad \text{(True)} \]
Since both original equations are satisfied, our solution is verified. Thus, the final solution is:
\[ \boxed{(8, 6)} \]