https://file.aiquickdraw.com/mathbot/file/1725904725685cn5sm26h.jpg

The system of linear equations shown in the image is:

\[
\begin{aligned}
1. \, x + y + z &= -1 \\
2. \, 3x + 5y + 4z &= 2 \\
3. \, 3x + 6y + 5z &= 0
\end{aligned}
\]

Let's solve this system using the method of elimination or substitution. I'll begin by solving step-by-step.

### Step 1: Eliminate one variable

We can subtract the second equation from the third to eliminate \( x \) (since the coefficients of \( x \) are the same).

\[
(3x + 6y + 5z) - (3x + 5y + 4z) = 0 - 2
\]

This simplifies to:

\[
y + z = -2 \quad \text{(Equation 4)}
\]

### Step 2: Substitution

From Equation 1, we know:

\[
x + y + z = -1 \quad \text{(Equation 1)}
\]

Substitute \( y + z = -2 \) (Equation 4) into Equation 1:

\[
x + (-2) = -1
\]

This simplifies to:

\[
x = 1
\]

### Step 3: Substitute \( x = 1 \) into the original equations

Now, substitute \( x = 1 \) into the second and third original equations.

For the second equation:

\[
3(1) + 5y + 4z = 2
\]

This simplifies to:

\[
3 + 5y + 4z = 2 \quad \Rightarrow \quad 5y + 4z = -1 \quad \text{(Equation 5)}
\]

For the third equation:

\[
3(1) + 6y + 5z = 0
\]

This simplifies to:

\[
3 + 6y + 5z = 0 \quad \Rightarrow \quad 6y + 5z = -3 \quad \text{(Equation 6)}
\]

### Step 4: Solve the system of two variables \( y \) and \( z \)

Now we solve the system:

\[
\begin{aligned}
5y + 4z &= -1 \quad \text{(Equation 5)} \\
6y + 5z &= -3 \quad \text{(Equation 6)}
\end{aligned}
\]

I'll solve this system for \( y \) and \( z \).The solution for the variables \( y \) and \( z \) is:

\[
y = 7, \quad z = -9
\]

### Step 5: Final solution

Now we have all the values for the variables:

\[
x = 1, \quad y = 7, \quad z = -9
\]

Thus, the solution to the system of equations is:

\[
(x, y, z) = (1, 7, -9)
\]

Would you like more details or have any further questions? Here are five related questions:

1. How can you verify the solution by substituting the values back into the original equations?
2. What is the geometric interpretation of solving a system of three equations in three variables?
3. Could you solve this system using matrix methods like Gaussian elimination?
4. How does substitution differ from elimination when solving systems of equations?
5. What happens if a system of linear equations has no solution or infinite solutions?

**Tip**: Always check your final solutions by substituting them back into the original equations to ensure they satisfy all the conditions.

Solve the system of linear equations: x + y + z = -1, 3x + 5y + 4z = 2, 3x + 6y + 5z = 0.

Learn how to solve the system of linear equations x + y + z = -1, 3x + 5y + 4z = 2, 3x + 6y + 5z = 0 using algebraic methods such as substitution and elimination. This problem involves key mathematical concepts suitable for students in Grades 9-12.

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