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

3x+4y=48

6x+2y=60

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

3. After you complete step 2, what would your top equation be?

4. Line them up and add the equations together to get this equation.

Solve for y.

Solve for x.

3 answers

To solve the given system of equations:

  1. The equations are: \[ 3x + 4y = 48 \quad \text{(Equation 1)} \] \[ 6x + 2y = 60 \quad \text{(Equation 2)} \]

  2. To cancel out the \(x\) variables, we can multiply Equation 1 by 2 so that the coefficients of \(x\) in both equations match. This gives us: \[ 2(3x + 4y) = 2(48) \implies 6x + 8y = 96 \quad \text{(New Equation 1)} \]

  3. The new top equation after multiplying Equation 1 by 2 is: \[ 6x + 8y = 96 \quad \text{(New Equation 1)} \]

  4. Now, we can line up the two equations: \[ 6x + 8y = 96 \quad \text{(New Equation 1)} \] \[ 6x + 2y = 60 \quad \text{(Equation 2)} \]

    Now, we subtract Equation 2 from New Equation 1 to eliminate \(6x\): \[ (6x + 8y) - (6x + 2y) = 96 - 60 \] Simplifying this gives: \[ 8y - 2y = 36 \implies 6y = 36 \]

    Now we can solve for \(y\): \[ y = \frac{36}{6} = 6 \]

  5. Now that we have \(y = 6\), we can substitute this value back into one of the original equations to solve for \(x\). We can use Equation 1: \[ 3x + 4(6) = 48 \] Simplifying this gives: \[ 3x + 24 = 48 \] \[ 3x = 48 - 24 \] \[ 3x = 24 \] \[ x = \frac{24}{3} = 8 \]

So the solution to the system of equations is: \[ x = 8, \quad y = 6 \]

sorry for the first one is this 1. Choose the best method to solve by looking at the way the equations are set up for you:

No problem! Let's choose the best method to solve the system of equations based on their setup. The equations are:

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

Best Method

The two equations have different coefficients in front of \(x\) and \(y\). To eliminate \(x\), we can use the elimination method since both equations can be manipulated to align nicely and easily cancel out one variable.

Steps for the Elimination Method

  1. Make the coefficients of \(x\) in both equations the same. Since Equation 1 already has a coefficient of \(3\) for \(x\), we can multiply the first equation by \(2\) to make the coefficient \(6\) (to match the coefficient of \(x\) in Equation 2): \[ 2(3x + 4y) = 2(48) \implies 6x + 8y = 96 \quad \text{(New Equation 1)} \]

  2. Now line up the equations: \[ 6x + 8y = 96 \quad \text{(New Equation 1)} \] \[ 6x + 2y = 60 \quad \text{(Equation 2)} \]

  3. Subtract Equation 2 from New Equation 1: \[ (6x + 8y) - (6x + 2y) = 96 - 60 \] This simplifies to: \[ 8y - 2y = 36 \] \[ 6y = 36 \] Now, solving for \(y\): \[ y = \frac{36}{6} = 6 \]

  4. Now substitute \(y\) back into one of the original equations (we can use Equation 1): \[ 3x + 4(6) = 48 \] This becomes: \[ 3x + 24 = 48 \] Now, isolating \(x\): \[ 3x = 48 - 24 \] \[ 3x = 24 \] \[ x = \frac{24}{3} = 8 \]

Final Solution

Thus, the solution to the system of equations is: \[ x = 8, \quad y = 6 \]

This method efficiently utilizes the structure of the equations to eliminate one variable, making the process straightforward.