4x1 + 2x2 − x3 = 2
3x1 − x2 + 2x3 = 10
11x1 + 3x2 = 8

We have the following system of linear equations:

1. \( 4x_1 + 2x_2 - x_3 = 2 \)
2. \( 3x_1 - x_2 + 2x_3 = 10 \)
3. \( 11x_1 + 3x_2 = 8 \)

Let’s solve this system using an algebraic method, such as substitution or elimination.

### Step 1: Solve Equation 3 for \(x_1\)
From equation (3):
\[
11x_1 + 3x_2 = 8
\]
Solve for \(x_1\):
\[
x_1 = \frac{8 - 3x_2}{11}
\]

### Step 2: Substitute \(x_1\) into the other equations
Substitute \(x_1 = \frac{8 - 3x_2}{11}\) into equations (1) and (2).

For equation (1):
\[
4\left(\frac{8 - 3x_2}{11}\right) + 2x_2 - x_3 = 2
\]
Simplify:
\[
\frac{32 - 12x_2}{11} + 2x_2 - x_3 = 2
\]
Multiply through by 11 to eliminate the denominator:
\[
32 - 12x_2 + 22x_2 - 11x_3 = 22
\]
Simplify:
\[
32 + 10x_2 - 11x_3 = 22
\]
\[
10x_2 - 11x_3 = -10 \quad \text{(Equation 4)}
\]

For equation (2):
\[
3\left(\frac{8 - 3x_2}{11}\right) - x_2 + 2x_3 = 10
\]
Simplify:
\[
\frac{24 - 9x_2}{11} - x_2 + 2x_3 = 10
\]
Multiply through by 11:
\[
24 - 9x_2 - 11x_2 + 22x_3 = 110
\]
Simplify:
\[
-20x_2 + 22x_3 = 86
\]
Divide by 2:
\[
-10x_2 + 11x_3 = 43 \quad \text{(Equation 5)}
\]

### Step 3: Solve the system of two equations
We now have the system:
1. \(10x_2 - 11x_3 = -10\)
2. \(-10x_2 + 11x_3 = 43\)

Add the two equations:
\[
(10x_2 - 11x_3) + (-10x_2 + 11x_3) = -10 + 43
\]
\[
0 = 33
\]

This is a contradiction, which means the system has **no solution**. The equations are inconsistent, indicating that the lines or planes represented by these

Explore how to solve a system of linear equations, including 4x1 + 2x2 − x3 = 2, using algebraic methods like substitution and elimination. The system has no solution, illustrating the concept of inconsistent systems. Suitable for Grades 10-12.

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