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

Question

3x+4y=48

6x+2y=60

options:

Choose the best method to solve by looking at the way the equations are set up for you:

Answer 1

To solve the given system of equations:

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

we can use either substitution or elimination. However, since the coefficients in the second equation can simplify more easily to match the first, let's consider a method that takes advantage of this.

Step 1: Simplify Equation 2

We can simplify Equation 2 by dividing all terms by 2:

\[ \frac{6x + 2y}{2} = \frac{60}{2} \implies 3x + y = 30 \]

So now, our system of equations becomes:

1. \(3x + 4y = 48\)
2. \(3x + y = 30\)

Step 2: Use Elimination

We can subtract the second equation from the first equation to eliminate \(3x\):

\[ (3x + 4y) - (3x + y) = 48 - 30 \]

This simplifies to:

\[ 3y = 18 \]

Step 3: Solve for \(y\)

Now, divide both sides by 3:

\[ y = 6 \]

Step 4: Solve for \(x\)

Now that we have \(y\), we can substitute it back into one of the original equations. Let's use Equation 2:

\[ 3x + y = 30 \implies 3x + 6 = 30 \]

Subtract 6 from both sides:

\[ 3x = 24 \]

Now, divide by 3:

\[ x = 8 \]

Solution

The solution to the system is:

\[ x = 8, \quad y = 6 \]

Conclusion

The best method to solve this system was elimination after simplifying the second equation.

Answer 2

Using the best method, how could you get the x variables to cancel out?

Answer 3

To eliminate the \(x\) variable from the given system of equations:

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

we can use the elimination method. To do this effectively, we need to manipulate the equations so that the coefficients of \(x\) in both equations are opposites (so they can cancel out). Here's how to proceed:

Step 1: Align the Coefficients of \(x\)

To make it easier to eliminate \(x\), we can multiply Equation 1 by 2, which will make the coefficient of \(x\) in both equations the same. This gives us:

\[ 2(3x + 4y) = 2(48) \implies 6x + 8y = 96 \quad \text{(New Equation 1)} \]

Now we have two equations:

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

Step 2: Subtract the Equations

Next, we can subtract Equation 2 from New Equation 1:

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

This simplifies to:

\[ 6y = 36 \]

Step 3: Solve for \(y\)

Now, divide both sides by 6:

\[ y = 6 \]

Step 4: Substitute to Find \(x\)

Now that we have \(y\), substitute it back into either original equation to solve for \(x\). Let’s use Equation 1:

\[ 3x + 4(6) = 48 \] \[ 3x + 24 = 48 \]

Subtract 24 from both sides:

\[ 3x = 24 \]

Now, divide by 3:

\[ x = 8 \]

Conclusion

The solution to the system remains:

\[ x = 8, \quad y = 6 \]

Summary

To eliminate the \(x\) variable, we multiplied the first equation by 2 to align the coefficients of \(x\) and then subtracted the equations. This allowed us to isolate and find the value of \(y\) first, and then use it to find \(x\).