Math Problem Statement

Solve the system of linear equations. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, express x1, x2, and x3 in terms of the parameter t.) 2x1 + x2 − 3x3 = 4 4x1 + 2x3 = 10 −6x1 + 3x2 − 15x3 = −18 (x1, x2, x3) =

Solution

We are given a system of three linear equations:

  1. 2x1+x23x3=42x_1 + x_2 - 3x_3 = 4
  2. 4x1+2x3=104x_1 + 2x_3 = 10
  3. 6x1+3x215x3=18-6x_1 + 3x_2 - 15x_3 = -18

We will solve this system step by step.

Step 1: Simplify the second equation

The second equation is: 4x1+2x3=104x_1 + 2x_3 = 10 We can divide the entire equation by 2: 2x1+x3=5(Equation 4)2x_1 + x_3 = 5 \quad \text{(Equation 4)}

Step 2: Solve the first equation for x2x_2

From the first equation: 2x1+x23x3=42x_1 + x_2 - 3x_3 = 4 We solve for x2x_2: x2=42x1+3x3(Equation 5)x_2 = 4 - 2x_1 + 3x_3 \quad \text{(Equation 5)}

Step 3: Simplify the third equation

The third equation is: 6x1+3x215x3=18-6x_1 + 3x_2 - 15x_3 = -18 We divide the entire equation by 3: 2x1+x25x3=6(Equation 6)-2x_1 + x_2 - 5x_3 = -6 \quad \text{(Equation 6)}

Step 4: Substitute x2x_2 from Equation 5 into Equation 6

Substitute x2=42x1+3x3x_2 = 4 - 2x_1 + 3x_3 into Equation 6: 2x1+(42x1+3x3)5x3=6-2x_1 + (4 - 2x_1 + 3x_3) - 5x_3 = -6 Simplifying: 2x1+42x1+3x35x3=6-2x_1 + 4 - 2x_1 + 3x_3 - 5x_3 = -6 Combine like terms: 4x12x3+4=6-4x_1 - 2x_3 + 4 = -6 Subtract 4 from both sides: 4x12x3=10-4x_1 - 2x_3 = -10 Divide by -2: 2x1+x3=5(Equation 7)2x_1 + x_3 = 5 \quad \text{(Equation 7)}

Step 5: Compare Equation 7 with Equation 4

Equation 7 is the same as Equation 4: 2x1+x3=52x_1 + x_3 = 5 This indicates that there are infinitely many solutions because the equations are dependent. We can express the solution in terms of a parameter tt.

Step 6: Express x1x_1, x2x_2, and x3x_3 in terms of tt

Let x3=tx_3 = t (parameter tt). Then from Equation 4: 2x1+t=52x_1 + t = 5 Solving for x1x_1: x1=5t2x_1 = \frac{5 - t}{2}

Substitute x1x_1 and x3x_3 into Equation 5 to solve for x2x_2: x2=42(5t2)+3tx_2 = 4 - 2\left(\frac{5 - t}{2}\right) + 3t Simplifying: x2=4(5t)+3tx_2 = 4 - (5 - t) + 3t x2=45+t+3tx_2 = 4 - 5 + t + 3t x2=1+4tx_2 = -1 + 4t

Final Answer:

(x1,x2,x3)=(5t2,4t1,t)(x_1, x_2, x_3) = \left(\frac{5 - t}{2}, 4t - 1, t\right)

This system has infinitely many solutions, with x1x_1, x2x_2, and x3x_3 expressed in terms of the parameter tt.

Would you like more details on any of the steps? Here are some related questions:

  1. What does it mean for a system of equations to have infinitely many solutions?
  2. How can we verify if a system of equations has no solution?
  3. What are some methods to solve systems of linear equations?
  4. How does the elimination method compare to substitution for solving such systems?
  5. What is the role of parameters in describing solution sets for systems with infinite solutions?

Tip: When a system has dependent equations, always try to simplify and identify relations between the variables using parameters!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Equations
Parametric Solutions

Formulas

Linear equation in three variables: ax1 + bx2 + cx3 = d
Solving dependent systems using parametric form

Theorems

Row Reduction Method
Consistency of Linear Systems
Dependent Systems

Suitable Grade Level

College Algebra or Advanced High School Algebra