using elimination solving systems of equations 3 variables 4a+5b-6c=2 -3a-2b+7c=-15 -a+4b+2c=-13


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.

using elimination solving systems of equations 3 variables 4a+5b-6c=2 -3a-2b+7c=-15 -a+4b+2c=-13

Learn how to solve a system of equations with three variables (4a + 5b - 6c = 2, -3a - 2b + 7c = -15, -a + 4b + 2c = -13) using the elimination method. Follow a detailed step-by-step process to eliminate variables and solve for a, b, and c. Suitable for students in Grades 10-12.

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