Math Problem Statement

x + y − 3z = −2 x − 2y + 6z = 10 −2x + y − 3z = −8

Solution

You are working with a system of three linear equations:

  1. x+y3z=2x + y - 3z = -2
  2. x2y+6z=10x - 2y + 6z = 10
  3. 2x+y3z=8-2x + y - 3z = -8

We can solve this system using substitution, elimination, or matrix methods like Gaussian elimination or Cramer's rule. Let's use elimination to solve it step-by-step.

Step 1: Simplify the system

Label the equations for convenience:

  • (1)x+y3z=2(1) \, x + y - 3z = -2
  • (2)x2y+6z=10(2) \, x - 2y + 6z = 10
  • (3)2x+y3z=8(3) \, -2x + y - 3z = -8

Step 2: Eliminate one variable

To eliminate xx, multiply equation (1) by 2 and add it to equation (3): 2(x+y3z)=2(2)    2x+2y6z=42(x + y - 3z) = 2(-2) \implies 2x + 2y - 6z = -4 Now add this to equation (3): (2x+y3z)+(2x+2y6z)=8+(4)(-2x + y - 3z) + (2x + 2y - 6z) = -8 + (-4) (0x)+3y9z=12    3y9z=12    y3z=4(4)(0x) + 3y - 9z = -12 \implies 3y - 9z = -12 \implies y - 3z = -4 \quad (4)

Step 3: Eliminate xx from another pair of equations

Subtract equation (1) from equation (2): (x2y+6z)(x+y3z)=10(2)(x - 2y + 6z) - (x + y - 3z) = 10 - (-2) 0x3y+9z=12    3y+9z=12    y3z=4(5)0x - 3y + 9z = 12 \implies -3y + 9z = 12 \implies y - 3z = -4 \quad (5)

Notice that equation (4) and equation (5) are identical, so no new information is gained from this step. This means the system has dependent equations and likely has infinitely many solutions.

Step 4: Express variables in terms of one free variable

From y3z=4y - 3z = -4, solve for yy: y=3z4y = 3z - 4 Substitute y=3z4y = 3z - 4 into equation (1) to solve for xx: x+(3z4)3z=2x + (3z - 4) - 3z = -2 Simplify: x4=2    x=2x - 4 = -2 \implies x = 2

Step 5: Final solution

Thus, the solution is: x=2,y=3z4x = 2, \quad y = 3z - 4 Here, zz is a free variable. The solution set is dependent on zz.

So, the general solution is: x=2,y=3z4,z=z(free variable)x = 2, \quad y = 3z - 4, \quad z = z \quad (\text{free variable})

Would you like more details or clarification on any part?


Here are 5 related questions to explore this topic further:

  1. How would the solution change if the system had a unique solution instead of infinitely many?
  2. How can you verify that the system has infinitely many solutions?
  3. Can you express the solution in vector form using parameterization?
  4. What are the conditions for a system of equations to have no solution?
  5. How does the determinant of the coefficient matrix relate to the number of solutions?

Tip: When solving a system of equations, checking for consistency between equations can help identify if the system has infinitely many or no solutions.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Systems of Equations

Formulas

Elimination method: Substituting one equation into another to eliminate a variable
General form of a linear equation: ax + by + cz = d

Theorems

Dependent system of equations theorem: Infinite solutions occur when two or more equations in the system are multiples of each other

Suitable Grade Level

Grades 9-12