To solve this system, you could do the following steps:

Question

3x+4y=48
6x+2y=60

Choose the best method to solve by looking at the way the equations are set up for you:
Using the best method, how could you get the x variables to cancel out?
After you complete step 2, what would your top equation be?
Line them up and add the equations together to get this equation.
Solve for y.
Solve for x.

Answer 1

To solve the system of equations:

1. \( 3x + 4y = 48 \) (Equation 1)
2. \( 6x + 2y = 60 \) (Equation 2)

Step 1: Choose the best method

A good method to solve this system is the elimination method. We can manipulate the equations to get the \( x \) variables to cancel each other out.

Step 2: Get the \( x \) variables to cancel out

To do this, we can multiply Equation 1 by 2 so that the coefficients of \( x \) will match:

\[ 2(3x + 4y) = 2(48) \] This simplifies to:

\[ 6x + 8y = 96 \quad \text{(New Equation 1)} \]

Now we have:

1. \( 6x + 8y = 96 \) (New Equation 1)
2. \( 6x + 2y = 60 \) (Equation 2)

Step 3: Line them up and subtract

Next, we subtract Equation 2 from the New Equation 1 to eliminate \( x \):

\[ (6x + 8y) - (6x + 2y) = 96 - 60 \]

This simplifies to:

\[ (8y - 2y) = 36 \] \[ 6y = 36 \]

Step 4: Solve for \( y \)

Now, solve for \( y \):

\[ y = \frac{36}{6} = 6 \]

Step 5: Solve for \( x \)

Now that we have \( y \), we can substitute \( y = 6 \) back into either original equation to solve for \( x \). We'll use Equation 1:

\[ 3x + 4(6) = 48 \] \[ 3x + 24 = 48 \] Subtract 24 from both sides:

\[ 3x = 48 - 24 \] \[ 3x = 24 \]

Divide by 3:

\[ x = \frac{24}{3} = 8 \]

Final solution

The solution to the system of equations is:

• \( x = 8 \)
• \( y = 6 \)