To solve the system of equations using the elimination method, we start with the two equations:
- \( -8x + 4y = 0 \)
- \( -x + 2y = -9 \)
Step 1: Align the equations for elimination.
We can make the second equation consistent with the first by multiplying it by 4, so that the coefficients of \(y\) match.
\[ 4(-x + 2y) = 4(-9) \]
This gives us:
\[ -4x + 8y = -36 \quad \text{(Equation 3)} \]
Now we have:
- \( -8x + 4y = 0 \)
- \( -4x + 8y = -36 \)
Step 2: Eliminate one variable.
We can multiply Equation 1 by 2 to line it up with Equation 3:
\[ 2(-8x + 4y) = 2(0) \] This simplifies to:
\[ -16x + 8y = 0 \quad \text{(Equation 4)} \]
Now we have:
- \( -16x + 8y = 0 \)
- \( -4x + 8y = -36 \)
Step 3: Subtract Equation 3 from Equation 4.
Now we can eliminate \(y\) by subtracting:
\[ (-16x + 8y) - (-4x + 8y) = 0 - (-36) \]
This simplifies to:
\[ -16x + 4x = 36 \] \[ -12x = 36 \]
Step 4: Solve for \(x\).
Divide both sides by \(-12\):
\[ x = \frac{36}{-12} = -3 \]
Step 5: Substitute \(x\) back to solve for \(y\).
Now we substitute \(x = -3\) back into one of the original equations. We'll use the second equation \( -x + 2y = -9 \):
\[ -(-3) + 2y = -9 \] \[ 3 + 2y = -9 \] \[ 2y = -9 - 3 \] \[ 2y = -12 \] \[ y = \frac{-12}{2} = -6 \]
Solution:
The solution to the system of equations is:
\[ x = -3, \quad y = -6 \]
Thus, the solution is \( (x, y) = (-3, -6) \).