We are given the following system of equations:

1. $y = -3x + 12$
2. $6y - x + 4 = 0$

Step 1: Substitute the expression for $y$ from Equation 1 into Equation 2.

From Equation 1, we know: $y = -3x + 12$

Now substitute this into Equation 2:

$6(-3x + 12) - x + 4 = 0$

Step 2: Simplify and solve for $x$.

Expand the terms: $6(-3x + 12) = -18x + 72$ Now substitute back: $-18x + 72 - x + 4 = 0$ Combine like terms: $-19x + 76 = 0$ Solve for $x$: $-19x = -76$ $x = 4$

Step 3: Solve for $y$.

Now that we have $x = 4$, substitute it back into Equation 1 to find $y$:

$y = -3(4) + 12 = -12 + 12 = 0$

Step 4: Check the solution.

Substitute $x = 4$ and $y = 0$ into the second equation:

$6(0) - 4 + 4 = 0$ $0 = 0$ Since the equation holds true, the solution is correct.

Final Answer:

The solution to the system of equations is $x = 4$ and $y = 0$.

Would you like more details or have any questions?

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Related Questions:

1. How does the substitution method work in solving systems of equations?
2. What other methods can be used to solve systems of linear equations?
3. How do you solve a system of equations graphically?
4. What happens when the system of equations has no solution?
5. Can the substitution method be applied to non-linear systems?

Tip: Always substitute your solution back into both original equations to ensure your solution is correct!