Let's solve the system of equations step by step using elimination or substitution.
The given equations are:
- \( 3x + 4y = 48 \) (Equation 1)
- \( 6x + 2y = 60 \) (Equation 2)
Step 1: Choose the best method to solve.
Response area: Elimination method
Step 2: How to get the x variables to cancel out?
To eliminate \(x\), we can multiply Equation 1 by 2, making the coefficients of \(x\) the same in both equations.
Response area: Multiply the first equation (Equation 1) by 2.
Step 3: What would your top equation be after step 2?
After multiplying Equation 1 by 2: \[ 2(3x + 4y) = 2(48) \implies 6x + 8y = 96 \] So, the new top equation is: Response area: \(6x + 8y = 96\)
Step 4: Line them up and add the equations together.
Now we have:
- \(6x + 8y = 96\) (Modified Equation 1)
- \(6x + 2y = 60\) (Equation 2)
Let's subtract Equation 2 from the modified Equation 1 to eliminate \(x\): \[ (6x + 8y) - (6x + 2y) = 96 - 60 \] This simplifies to: \[ 6y = 36 \]
Response area: \(6y = 36\)
Step 5: Solve for y.
To find \(y\), divide both sides by 6: \[ y = \frac{36}{6} = 6 \]
Response area: \(y = 6\)
Step 6: Solve for x.
Now that we have \(y\), we can substitute it back into one of the original equations to solve for \(x\). Let's use Equation 1: \[ 3x + 4(6) = 48 \] This simplifies to: \[ 3x + 24 = 48 \implies 3x = 48 - 24 \implies 3x = 24 \] Dividing by 3 gives us: \[ x = \frac{24}{3} = 8 \]
Response area: \(x = 8\)
Final solution:
- \(x = 8\)
- \(y = 6\)
Feel free to use this format in your input fields as needed.