To solve the system of equations:
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Choose the best method to solve by looking at the way the equations are set up for you:
- The equations are easy to manipulate, so the elimination method (also known as the addition method) is a good choice since it allows for straightforward cancellation of one of the variables.
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Using the best method, how could you get the x variables to cancel out?
- To eliminate \( x \), we need to make the coefficients of \( x \) in both equations equal. We can multiply the first equation by 2, which will transform \( 3x \) into \( 6x \), matching the coefficient of \( x \) in the second equation.
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After you complete step 2, what would your top equation be?
- The first equation, \( 3x + 4y = 48 \), multiplied by 2 becomes: \[ 6x + 8y = 96 \]
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Line them up and add the equations together to get this equation.
- Now we have: \[ 6x + 8y = 96 \quad \text{(new first equation)} \] \[ 6x + 2y = 60 \quad \text{(second equation)} \]
- Next, we will subtract the second equation from the new first equation: \[ (6x + 8y) - (6x + 2y) = 96 - 60 \]
- This simplifies to: \[ 6y = 36 \]
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Solve for y.
- Divide both sides by 6: \[ y = 6 \]
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Solve for x.
- Now substitute \( y = 6 \) back into one of the original equations. We'll use the second equation \( 6x + 2y = 60 \): \[ 6x + 2(6) = 60 \] \[ 6x + 12 = 60 \] \[ 6x = 60 - 12 \] \[ 6x = 48 \] \[ x = 8 \]
The solution to the system of equations is: \[ x = 8, \quad y = 6 \]
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