To solve the system of equations using the substitution method, let's first observe the system of equations:
- \( 2x - y + 3z = 3 \) (Equation 1)
- \( 4x - 2y + 6z = 6 \) (Equation 2)
- \( -6x + 3y - 9z = -9 \) (Equation 3)
Step 1: Simplify the equations.
Let's simplify Equation 2 and Equation 3 to see if they can be derived from Equation 1.
- For Equation 2, if we divide everything by 2:
\[ 4x - 2y + 6z = 6 \implies 2x - y + 3z = 3 \quad \text{(This is exactly Equation 1)} \]
- For Equation 3, if we divide everything by -3:
\[ -6x + 3y - 9z = -9 \implies 2x - y + 3z = 3 \quad \text{(This is again Equation 1)} \]
Step 2: Analyze the results.
Thus, we see that Equations 2 and 3 are merely scalar multiples of Equation 1. This indicates that all three equations represent the same plane in three-dimensional space, meaning they are dependent.
Since there are infinitely many solutions, we can express \(y\) and \(z\) in terms of \(x\).
Step 3: Solve for \(y\) and \(z\) in terms of \(x\).
From Equation 1, we can express \(y\):
\[ 2x - y + 3z = 3 \implies y = 2x + 3z - 3 \]
Step 4: Parameterize the solution.
If we let \(z\) be a parameter (denoted as \(t\)), then we can write:
\[ y = 2x + 3t - 3 \]
Where \(x\) can take any real number value \(k\). So, we can express the solution in terms of \(x\) and \(z\):
\[ x = k, \quad y = 2k + 3t - 3, \quad z = t \]
Thus, any value of \(x\) will yield corresponding dependent values for \(y\) and \(z\).
Final Conclusion:
Since the original question asks only for the value of \(x\), and no specific value was provided for \(x\) in the equations, we conclude that \(x\) can take any value.
Thus, the response to the question regarding the value of \(x\) is that there are infinitely many solutions, and \(x\) can be any real number.
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