Math Problem Statement

Solve the System. Give answer as (x,y,z).

{-2x-y-5z=19. -4x+y+2z=-10. 8x+y-6z=14

(x,y,z)=

Solution

We are given the system of equations:

-2x - y - 5z &= 19 \quad \text{(Equation 1)} \\ -4x + y + 2z &= -10 \quad \text{(Equation 2)} \\ 8x + y - 6z &= 14 \quad \text{(Equation 3)} \end{aligned}$$ ### Step 1: Add Equation 1 and Equation 2 to eliminate $$y$$ $$(-2x - y - 5z) + (-4x + y + 2z) = 19 + (-10)$$ This simplifies to: $$-6x - 3z = 9 \quad \Rightarrow \quad -2x - z = 3 \quad \text{(Equation 4)}$$ ### Step 2: Add Equation 2 and Equation 3 to eliminate $$y$$ $$(-4x + y + 2z) + (8x + y - 6z) = -10 + 14$$ This simplifies to: $$4x - 4z = 4 \quad \Rightarrow \quad x - z = 1 \quad \text{(Equation 5)}$$ ### Step 3: Solve the system of two equations in $$x$$ and $$z$$ We now have: $$-2x - z = 3 \quad \text{(Equation 4)}$$ $$x - z = 1 \quad \text{(Equation 5)}$$ From Equation 5, solve for $$z$$: $$z = x - 1$$ Substitute this into Equation 4: $$-2x - (x - 1) = 3$$ Simplify: $$-2x - x + 1 = 3$$ $$-3x + 1 = 3$$ $$-3x = 2 \quad \Rightarrow \quad x = -\frac{2}{3}$$ ### Step 4: Find $$z$$ Substitute $$x = -\frac{2}{3}$$ into $$z = x - 1$$: $$z = -\frac{2}{3} - 1 = -\frac{2}{3} - \frac{3}{3} = -\frac{5}{3}$$ ### Step 5: Find $$y$$ Substitute $$x = -\frac{2}{3}$$ and $$z = -\frac{5}{3}$$ into any original equation, say Equation 1: $$-2(-\frac{2}{3}) - y - 5(-\frac{5}{3}) = 19$$ Simplify: $$\frac{4}{3} - y + \frac{25}{3} = 19$$ $$\frac{29}{3} - y = 19$$ Multiply both sides by 3: $$29 - 3y = 57$$ $$-3y = 57 - 29 = 28$$ $$y = -\frac{28}{3}$$ ### Final Answer: The solution to the system is: $$(x, y, z) = \left( -\frac{2}{3}, -\frac{28}{3}, -\frac{5}{3} \right)$$ Would you like more details or have any questions? Here are five related questions to deepen understanding: 1. How do we verify that the solution satisfies all three original equations? 2. What other methods could be used to solve this system of equations? 3. How can substitution be used more efficiently in systems with more equations? 4. How would the solution change if one of the equations were modified? 5. What is the geometric interpretation of solving systems of three equations? **Tip**: When solving systems of equations, substitution and elimination methods are both useful. Choosing the method depends on how easily variables can be isolated.

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Equations
Linear Equations

Formulas

Substitution method
Elimination method

Theorems

Linear Combination
System of Linear Equations

Suitable Grade Level

Grades 10-12

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