To solve the system of equations using the elimination method, we'll start with the given equations:
- \( \frac{1}{3}x + y = 3 \)
- \( 2y = -\frac{2}{3}x + 10 \)
Let's first eliminate the fractions by multiplying both equations to make them easier to work with.
Step 1: Multiply the equations to eliminate fractions
Equation 1: \[ \frac{1}{3}x + y = 3 \] Multiply by 3: \[ x + 3y = 9 \quad \text{(Equation 1')} \]
Equation 2: \[ 2y = -\frac{2}{3}x + 10 \] Multiply by 3: \[ 6y = -2x + 30 \quad \text{(Equation 2')} \]
Now we have a new system of equations:
- \( x + 3y = 9 \) (Equation 1')
- \( 2x + 6y = 30 \) (Equation 2')
Step 2: Align equations for elimination
We'll manipulate Equation 1' so we can eliminate \( y \) by aligning it with Equation 2'.
We can rewrite Equation 1' as: \[ 2(x + 3y) = 2(9) \] This gives: \[ 2x + 6y = 18 \quad \text{(Equation 3)} \]
Now we have:
- \( 2x + 6y = 18 \) (Equation 3)
- \( 2x + 6y = 30 \) (Equation 2')
Step 3: Subtract the equations
Now we can subtract Equation 3 from Equation 2': \[ (2x + 6y) - (2x + 6y) = 30 - 18 \] This simplifies to: \[ 0 = 12 \]
Step 4: Conclusion
Since we came to a false statement, this means that the two equations represent parallel lines and there is no solution to this system of equations. Thus, the system of equations is inconsistent.
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