To solve the system of equations with three variables using the elimination method, we begin with the following system of equations:

1. $4a + 5b - 6c = 2$
2. $-3a - 2b + 7c = -15$
3. $-a + 4b + 2c = -13$

Step 1: Eliminate one variable from two pairs of equations

We aim to eliminate one of the variables by adding or subtracting equations. Let's first try to eliminate $a$.

To eliminate $a$, we need to manipulate the coefficients of $a$ in each equation. We will use equations (1) and (2) first:

Multiply equation (2) by 4 and equation (1) by 3, so the coefficients of $a$ will cancel out:

$\text{(1)} \times 3: \quad 12a + 15b - 18c = 6$ $\text{(2)} \times 4: \quad -12a - 8b + 28c = -60$

Now, add these two equations to eliminate $a$:

$(12a + 15b - 18c) + (-12a - 8b + 28c) = 6 + (-60)$ $(15b - 8b) + (-18c + 28c) = -54$ $7b + 10c = -54 \quad \text{(Equation 4)}$

Step 2: Eliminate $a$ again using a different pair of equations

Now, we will eliminate $a$ using equations (1) and (3). First, we multiply equation (3) by 4 so that the coefficient of $a$ matches that of equation (1):

$\text{(1)} \times 1: \quad 4a + 5b - 6c = 2$ $\text{(3)} \times 4: \quad -4a + 16b + 8c = -52$

Now, add these two equations to eliminate $a$:

$(4a + 5b - 6c) + (-4a + 16b + 8c) = 2 + (-52)$ $(5b + 16b) + (-6c + 8c) = -50$ $21b + 2c = -50 \quad \text{(Equation 5)}$

Step 3: Solve the system of two equations with two variables

Now, we have the system of two equations:

1. $7b + 10c = -54$ (Equation 4)
2. $21b + 2c = -50$ (Equation 5)

Multiply equation (5) by 7 to align the coefficient of $b$:

$\text{(5)} \times 7: \quad 147b + 14c = -350$

Multiply equation (4) by 21 to align the coefficient of $b$:

$\text{(4)} \times 21: \quad 147b + 210c = -1134$

Subtract the two equations to eliminate $b$:

$(147b + 210c) - (147b + 14c) = -1134 - (-350)$ $196c = -784$ $c = \frac{-784}{196} = -4$

Step 4: Substitute $c = -4$ into one of the two-variable equations

Substitute $c = -4$ into equation (4):

$7b + 10(-4) = -54$ $7b - 40 = -54$ $7b = -54 + 40$ $7b = -14$ $b = \frac{-14}{7} = -2$

Step 5: Substitute $b = -2$ and $c = -4$ into one of the original equations

Now that we have $b = -2$ and $c = -4$, substitute these values into equation (1) to solve for $a$:

$4a + 5(-2) - 6(-4) = 2$ $4a - 10 + 24 = 2$ $4a + 14 = 2$ $4a = 2 - 14$ $4a = -12$ $a = \frac{-12}{4} = -3$

Final Solution:

The solution to the system of equations is: $a = -3, \quad b = -2, \quad c = -4$

Would you like any further details or clarifications? Here are some related questions you can explore:

1. How is substitution different from elimination in solving systems of equations?
2. Can elimination be used for non-linear systems?
3. How do you solve systems of equations in matrix form?
4. What are the real-world applications of solving systems of equations?
5. What happens if a system of equations has no solutions or infinitely many solutions?

Tip: When using elimination, carefully align variables and coefficients to avoid errors in calculation.