Math Problem Statement

–x+y–2z=5 –x+y+z=5 x+3y+z=–9

Solution

Let's solve the system of linear equations:

1) & \ -x + y - 2z = 5 \\ 2) & \ -x + y + z = 5 \\ 3) & \ x + 3y + z = -9 \end{aligned}$$ ### Step 1: Eliminate one variable We'll eliminate $$x$$ by adding equation 1 and equation 2: $$\begin{aligned} (-x + y - 2z) + (-x + y + z) &= 5 + 5 \\ -2x + 2y - z &= 10 \\ \text{Divide the entire equation by 2:} \\ -x + y - \frac{z}{2} &= 5 \end{aligned}$$ But we can simplify further if we solve stepwise: $$\text{Let's solve equation 1:} \\ (-x + y - 2z) - (-x + y + z) = 5 - 5 \quad \text{to eliminate} \, x \, \text{and} \, y: \\ -3z = 0 \quad \Rightarrow \quad z = 0$$ ### Step 2: Substitute $$z = 0$$ into equations 2 and 3 Substitute $$z = 0$$ into equation 2: $$-x + y + 0 = 5 \quad \Rightarrow \quad -x + y = 5$$ Now substitute $$z = 0$$ into equation 3: $$x + 3y + 0 = -9 \quad \Rightarrow \quad x + 3y = -9$$ ### Step 3: Solve for $$x$$ and $$y$$ Now we have two equations: $$1) \ -x + y = 5 \\ 2) \ x + 3y = -9$$ Add the two equations to eliminate $$x$$: $$(-x + y) + (x + 3y) = 5 - 9 \quad \Rightarrow \quad 4y = -4 \quad \Rightarrow \quad y = -1$$ Now, substitute $$y = -1$$ back into the equation $$-x + y = 5$$: $$-x - 1 = 5 \quad \Rightarrow \quad -x = 6 \quad \Rightarrow \quad x = -6$$ ### Final Solution The solution to the system is: $$x = -6, \quad y = -1, \quad z = 0$$ Would you like any further details or have any questions about the process? --- Here are some related questions you might find interesting: 1. How can we solve a system of linear equations using the substitution method? 2. What is the difference between the substitution method and the elimination method for solving systems of equations? 3. How can we determine if a system of equations has no solution, one solution, or infinitely many solutions? 4. How can matrices be used to solve a system of linear equations? 5. What is the geometric interpretation of the solution to a system of linear equations in three variables? **Tip:** When solving systems of equations, always double-check your solution by substituting it back into the original equations.