https://file.aiquickdraw.com/mathbot/file/17290576732363vjgapd1.jpg

The system of equations you provided from the image is:

\[
\begin{align*}
x_2 + x_3 + x_4 &= 0 \quad \text{(1)} \\
3x_1 + 3x_3 - 4x_4 &= 7 \quad \text{(2)} \\
x_1 + x_2 + x_3 + 2x_4 &= 6 \quad \text{(3)} \\
2x_1 + 3x_2 + x_3 + 3x_4 &= 6 \quad \text{(4)}
\end{align*}
\]

This is a system of four equations with four unknowns: \(x_1, x_2, x_3, x_4\).

I will now solve this system step by step using substitution or elimination methods.

### Step 1: Solve Equation (1)
From equation (1), we have:

\[
x_2 + x_3 + x_4 = 0 \quad \Rightarrow \quad x_2 = -x_3 - x_4
\]

### Step 2: Substitute \(x_2\) into the other equations
Substitute \(x_2 = -x_3 - x_4\) into equations (3) and (4):

**Substitute into equation (3):**

\[
x_1 + (-x_3 - x_4) + x_3 + 2x_4 = 6
\]

Simplifying:

\[
x_1 - x_4 + 2x_4 = 6 \quad \Rightarrow \quad x_1 + x_4 = 6 \quad \Rightarrow \quad x_1 = 6 - x_4 \quad \text{(5)}
\]

**Substitute into equation (4):**

\[
2x_1 + 3(-x_3 - x_4) + x_3 + 3x_4 = 6
\]

Simplifying:

\[
2x_1 - 3x_3 - 3x_4 + x_3 + 3x_4 = 6 \quad \Rightarrow \quad 2x_1 - 2x_3 = 6 \quad \Rightarrow \quad x_1 = 3 + x_3 \quad \text{(6)}
\]

### Step 3: Solve for \(x_1\), \(x_3\), and \(x_4\)
Now, using equations (5) and (6):

From equation (5): \(x_1 = 6 - x_4\)  
From equation (6): \(x_1 = 3 + x_3\)

Equating the two expressions for \(x_1\):

\[
6 - x_4 = 3 + x_3 \quad \Rightarrow \quad x_3 + x_4 = 3 \quad \text{(7)}
\]

Now substitute equation (7) into equation (2):

\[
3x_1 + 3x_3 - 4x_4 = 7
\]

From equation (5), \(x_1 = 6 - x_4\), so:

\[
3(6 - x_4) + 3x_3 - 4x_4 = 7
\]

Expanding:

\[
18 - 3x_4 + 3x_3 - 4x_4 = 7 \quad \Rightarrow \quad 18 + 3(x_3 - x_4) - 4x_4 = 7
\]

Now substitute equation (7) \(x_3 = 3 - x_4\) into the above:

\[
18 + 3(3 - x_4) - 4x_4 = 7
\]

Simplifying:

\[
18 + 9 - 3x_4 - 4x_4 = 7 \quad \Rightarrow \quad 27 - 7x_4 = 7
\]

Solve for \(x_4\):

\[
27 - 7x_4 = 7 \quad \Rightarrow \quad -7x_4 = -20 \quad \Rightarrow \quad x_4 = \frac{20}{7}
\]

### Step 4: Solve for \(x_1, x_2, x_3\)
Now that we know \(x_4 = \frac{20}{7}\), substitute it into the other equations:

From equation (5), \(x_1 = 6 - x_4\):

\[
x_1 = 6 - \frac{20}{7} = \frac{42}{7} - \frac{20}{7} = \frac{22}{7}
\]

From equation (7), \(x_3 + x_4 = 3\):

\[
x_3 = 3 - \frac{20}{7} = \frac{21}{7} - \frac{20}{7} = \frac{1}{7}
\]

Finally, from equation (1), \(x_2 + x_3 + x_4 = 0\):

\[
x_2 + \frac{1}{7} + \frac{20}{7} = 0 \quad \Rightarrow \quad x_2 + \frac{21}{7} = 0 \quad \Rightarrow \quad x_2 = -\frac{21}{7} = -3
\]

### Solution
The solution to the system of equations is:

\[
x_1 = \frac{22}{7}, \quad x_2 = -3, \quad x_3 = \frac{1}{7}, \quad x_4 = \frac{20}{7}
\]

Would you like more details or have any questions?

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Here are 5 related questions to explore further:
1. How do you solve a system of linear equations using matrix methods?
2. What are the conditions for a system of equations to have a unique solution?
3. Can the substitution method always be used for solving systems of equations?
4. What is the role of Gaussian elimination in solving linear systems?
5. How do underdetermined or overdetermined systems differ in their solutions?

**Tip**: When solving systems of equations, always check your solutions by substituting the values back into the original equations.

Solve the system of linear equations: x2 + x3 + x4 = 0, 3x1 + 3x3 - 4x4 = 7, x1 + x2 + x3 + 2x4 = 6, 2x1 + 3x2 + x3 + 3x4 = 6.

Learn how to solve a system of four linear equations using the substitution method. The equations involve variables x1, x2, x3, and x4, and the solution requires step-by-step elimination and substitution of variables. Ideal for students in Grades 9-12.

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