Math Problem Statement

x − y + z = 1 Equation 1 2x − y − 2z = 3 Equation 2 x + 3y − 3z = −1 solve using elimination

Solution

We are given the system of equations:

Equation 1: xy+z=1\text{Equation 1: } x - y + z = 1 Equation 2: 2xy2z=3\text{Equation 2: } 2x - y - 2z = 3 Equation 3: x+3y3z=1\text{Equation 3: } x + 3y - 3z = -1

Step 1: Eliminate one variable (let's eliminate yy)

First, subtract Equation 1 from Equation 2 to eliminate yy:

(2xy2z)(xy+z)=31(2x - y - 2z) - (x - y + z) = 3 - 1 Simplifying: 2xy2zx+yz=22x - y - 2z - x + y - z = 2 x3z=2(Equation 4)x - 3z = 2 \quad \text{(Equation 4)}

Next, subtract Equation 1 from Equation 3 to eliminate yy:

(x+3y3z)(xy+z)=11(x + 3y - 3z) - (x - y + z) = -1 - 1 Simplifying: x+3y3zx+yz=2x + 3y - 3z - x + y - z = -2 4y4z=24y - 4z = -2 Dividing by 2: 2y2z=1(Equation 5)2y - 2z = -1 \quad \text{(Equation 5)}

Step 2: Solve the system of two equations

Now we have the reduced system:

  1. x3z=2x - 3z = 2 (Equation 4)
  2. 2y2z=12y - 2z = -1 (Equation 5)

Solve for xx in terms of zz from Equation 4:

x=3z+2x = 3z + 2

Solve for yy in terms of zz from Equation 5:

2y=2z12y = 2z - 1 y=z12y = z - \frac{1}{2}

Step 3: Substitute into the original equation

Now that we have x=3z+2x = 3z + 2 and y=z12y = z - \frac{1}{2}, substitute these into Equation 1 to find zz:

xy+z=1x - y + z = 1 (3z+2)(z12)+z=1(3z + 2) - \left(z - \frac{1}{2}\right) + z = 1 Simplifying: 3z+2z+12+z=13z + 2 - z + \frac{1}{2} + z = 1 3z+52=13z + \frac{5}{2} = 1 3z=1523z = 1 - \frac{5}{2} 3z=323z = -\frac{3}{2} z=12z = -\frac{1}{2}

Step 4: Solve for xx and yy

Now that we know z=12z = -\frac{1}{2}, we can substitute this value into the expressions for xx and yy:

x=3z+2=3(12)+2=32+2=12x = 3z + 2 = 3\left(-\frac{1}{2}\right) + 2 = -\frac{3}{2} + 2 = \frac{1}{2} y=z12=1212=1y = z - \frac{1}{2} = -\frac{1}{2} - \frac{1}{2} = -1

Final Solution:

x=12,y=1,z=12x = \frac{1}{2}, \quad y = -1, \quad z = -\frac{1}{2}

Would you like further details or have any questions?

Here are 5 related questions to explore:

  1. How can substitution be used to solve this system?
  2. What happens if you choose to eliminate zz instead of yy?
  3. How does elimination compare to solving systems graphically?
  4. Can this method be applied to larger systems of equations?
  5. What real-world problems can be modeled by systems of equations?

Tip: Always double-check each step of elimination to avoid sign errors!

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Systems of Equations
Elimination Method

Formulas

Linear Equation: ax + by + cz = d
Elimination of variables by subtraction

Theorems

Elimination Method for Systems of Equations

Suitable Grade Level

Grades 9-11