Question

question 33 points 2

solve:
\\
\

$$\begin{aligned} 2x - y + z &= 7 \\\\ 3x + y - 5z &= 13 \\\\ x + y + z &= 5 \\end{aligned}$$

\\

\\(x = 1, y = 1, z = 0\\)

\\(x = 4, y = 1, z = 0\\)

Explanation:

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<reasoning>

Set up the system of equations

Using the Coefficient Matrix knowledge point
\[

$$\begin{aligned} 2x - y + z &= 7 \quad \text{(Equation 1)} \\ 3x + y - 5z &= 13 \quad \text{(Equation 2)} \\ x + y + z &= 5 \quad \text{(Equation 3)} \end{aligned}$$

\]

Eliminate variables to solve the system

We can eliminate \(y\) by adding Equation 1 and Equation 3:
\[

$$\begin{aligned} (2x - y + z) + (x + y + z) &= 7 + 5 \\ 3x + 2z &= 12 \quad \text{(Equation 4)} \end{aligned}$$

\]

Next, we eliminate \(y\) by subtracting Equation 3 from Equation 2:
\[

$$\begin{aligned} (3x + y - 5z) - (x + y + z) &= 13 - 5 \\ 2x - 6z &= 8 \\ x - 3z &= 4 \quad \text{(Equation 5)} \end{aligned}$$

\]

Solve the simplified system for x and z

From Equation 5, we express \(x\) in terms of \(z\):
\[
x = 3z + 4
\]

Substitute this expression for \(x\) into Equation 4:
\[

$$\begin{aligned} 3(3z + 4) + 2z &= 12 \\ 9z + 12 + 2z &= 12 \\ 11z &= 0 \\ z &= 0 \end{aligned}$$

\]

Substitute \(z = 0\) back to find \(x\):
\[
x = 3(0) + 4 = 4
\]

Find the value of y

Substitute \(x = 4\) and \(z = 0\) into Equation 3:
\[

$$\begin{aligned} 4 + y + 0 &= 5 \\ y &= 1 \end{aligned}$$

\]

Verify the solution and match options

Using the Matrix Determinant knowledge point
\[

$$\begin{aligned} \text{Check Equation 1: } 2(4) - 1 + 0 &= 7 \quad (\text{True}) \\ \text{Check Equation 2: } 3(4) + 1 - 5(0) &= 13 \quad (\text{True}) \\ \text{The solution is: } x = 4, y = 1, z = 0 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>\(x = 1, y = 1, z = 0\)</mcq-option>
<mcq-correct>\(x = 4, y = 1, z = 0\)</mcq-correct>
</answer>

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Answer:

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<reasoning>

Set up the system of equations

Using the Coefficient Matrix knowledge point
\[

$$\begin{aligned} 2x - y + z &= 7 \quad \text{(Equation 1)} \\ 3x + y - 5z &= 13 \quad \text{(Equation 2)} \\ x + y + z &= 5 \quad \text{(Equation 3)} \end{aligned}$$

\]

Eliminate variables to solve the system

We can eliminate \(y\) by adding Equation 1 and Equation 3:
\[

$$\begin{aligned} (2x - y + z) + (x + y + z) &= 7 + 5 \\ 3x + 2z &= 12 \quad \text{(Equation 4)} \end{aligned}$$

\]

Next, we eliminate \(y\) by subtracting Equation 3 from Equation 2:
\[

$$\begin{aligned} (3x + y - 5z) - (x + y + z) &= 13 - 5 \\ 2x - 6z &= 8 \\ x - 3z &= 4 \quad \text{(Equation 5)} \end{aligned}$$

\]

Solve the simplified system for x and z

From Equation 5, we express \(x\) in terms of \(z\):
\[
x = 3z + 4
\]

Substitute this expression for \(x\) into Equation 4:
\[

$$\begin{aligned} 3(3z + 4) + 2z &= 12 \\ 9z + 12 + 2z &= 12 \\ 11z &= 0 \\ z &= 0 \end{aligned}$$

\]

Substitute \(z = 0\) back to find \(x\):
\[
x = 3(0) + 4 = 4
\]

Find the value of y

Substitute \(x = 4\) and \(z = 0\) into Equation 3:
\[

$$\begin{aligned} 4 + y + 0 &= 5 \\ y &= 1 \end{aligned}$$

\]

Verify the solution and match options

Using the Matrix Determinant knowledge point
\[

$$\begin{aligned} \text{Check Equation 1: } 2(4) - 1 + 0 &= 7 \quad (\text{True}) \\ \text{Check Equation 2: } 3(4) + 1 - 5(0) &= 13 \quad (\text{True}) \\ \text{The solution is: } x = 4, y = 1, z = 0 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>\(x = 1, y = 1, z = 0\)</mcq-option>
<mcq-correct>\(x = 4, y = 1, z = 0\)</mcq-correct>
</answer>

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