Math Problem Statement
SPLTV<div>2x+y+z=2</div><div>x-y-2z=5</div><div>x+2y+z=9</div>
Solution
To solve the system of linear equations:
1. & \ \ 2x + y + z = 2 \\ 2. & \ \ x - y - 2z = 5 \\ 3. & \ \ x + 2y + z = 9 \end{align*}$$ We can use methods such as substitution, elimination, or matrix-based approaches like Gaussian elimination. I'll use elimination to solve the system step by step. ### Step 1: Eliminate one variable First, let's subtract equation 2 from equation 1 to eliminate $$y$$. $$(2x + y + z) - (x - y - 2z) = 2 - 5$$ Simplifying: $$x + 2y + 3z = -3$$ We now have a simpler system of equations: $$\begin{align*} 4. & \ \ x + 2y + z = 9 \quad \text{(equation 3)} \\ 5. & \ \ x + 2y + 3z = -3 \end{align*}$$ ### Step 2: Eliminate $$x$$ Next, subtract equation 4 from equation 5: $$(x + 2y + 3z) - (x + 2y + z) = -3 - 9$$ Simplifying: $$2z = -12 \quad \Rightarrow \quad z = -6$$ ### Step 3: Substitute $$z = -6$$ into one of the simpler equations Substitute $$z = -6$$ into equation 4: $$x + 2y + (-6) = 9$$ Simplifying: $$x + 2y - 6 = 9 \quad \Rightarrow \quad x + 2y = 15 \quad \text{(equation 6)}$$ ### Step 4: Use substitution to find $$x$$ and $$y$$ Now substitute $$z = -6$$ into equation 1: $$2x + y + (-6) = 2$$ Simplifying: $$2x + y - 6 = 2 \quad \Rightarrow \quad 2x + y = 8 \quad \text{(equation 7)}$$ We now have two equations with $$x$$ and $$y$$: $$\begin{align*} 6. & \ \ x + 2y = 15 \\ 7. & \ \ 2x + y = 8 \end{align*}$$ ### Step 5: Solve for $$x$$ and $$y$$ Multiply equation 6 by 2: $$2x + 4y = 30 \quad \text{(equation 8)}$$ Subtract equation 7 from equation 8: $$(2x + 4y) - (2x + y) = 30 - 8$$ Simplifying: $$3y = 22 \quad \Rightarrow \quad y = \frac{22}{3}$$ Now substitute $$y = \frac{22}{3}$$ into equation 6: $$x + 2 \times \frac{22}{3} = 15$$ Simplifying: $$x + \frac{44}{3} = 15 \quad \Rightarrow \quad x = 15 - \frac{44}{3} = \frac{45}{3} - \frac{44}{3} = \frac{1}{3}$$ ### Final Solution: $$x = \frac{1}{3}, \quad y = \frac{22}{3}, \quad z = -6$$ Would you like further explanation or a verification of the solution? Here are some related questions you might find useful: 1. How can I verify the solution using substitution? 2. What are some other methods to solve this system of equations? 3. How would the solution change if one of the constants in the system was different? 4. Can this system be solved using matrices or determinants? 5. How would I interpret this solution in a real-world context? **Tip:** Always check your solution by substituting the values back into the original equations to ensure accuracy.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Linear Equations
Elimination Method
Formulas
Elimination method to solve linear equations
Substitution for variable elimination
Matrix operations for solving systems of equations (optional approach)
Theorems
Solution of linear systems (consistent, inconsistent, dependent)
Suitable Grade Level
Grades 9-12
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