text: ## Elementary Linear Algebra 7th Edition
  
  Published by Cengage Learning
  
  # Chapter 1 - Systems of Linear Equations - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises - Page 10: 41
  
  #### Answer
  
  Ans: $x=-\frac{1}{3}, y=-\frac{2}{3}$
  
  #### Work Step by Step
  
  Step 1: Simplify second equation $\frac{1}{5}x+\frac{2}{5}y=-\frac{1}{3}$ $3x+6y=-5$ $6y=-5-3x$ Step 2: Put $6y$ to one side on first equation $3y=9x+1$ $6y=18x+2$ Step 3: The other side of $6y$ are equal, so make an equation and solve $x$ $18x+2=-5-3x$ $21x=-7$ $x=-\frac{1}{3}$ Step 4: Substitute $x$ to one of the equations $9\times(-\frac{1}{3})-3y=-1$ $-3y=2$ $y=-\frac{2}{3}$
  
  After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
explanation: The document is an excerpt from a textbook, "Elementary Linear Algebra 7th Edition," published by Cengage Learning. It contains a math problem from Chapter 1, Section 1.1, Exercise 41, which involves solving a system of linear equations. The excerpt includes the problem statement, the answer, and a step-by-step solution.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ## Elementary Linear Algebra 7th Edition
  
  Published by Cengage Learning
  
  # Chapter 1 - Systems of Linear Equations - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises - Page 10: 41
  
  #### Answer
  
  Ans: $x=-\frac{1}{3}, y=-\frac{2}{3}$
  
  #### Work Step by Step
  
  Step 1: Simplify second equation $\frac{1}{5}x+\frac{2}{5}y=-\frac{1}{3}$ $3x+6y=-5$ $6y=-5-3x$ Step 2: Put $6y$ to one side on first equation $3y=9x+1$ $6y=18x+2$ Step 3: The other side of $6y$ are equal, so make an equation and solve $x$ $18x+2=-5-3x$ $21x=-7$ $x=-\frac{1}{3}$ Step 4: Substitute $x$ to one of the equations $9\times(-\frac{1}{3})-3y=-1$ $-3y=2$ $y=-\frac{2}{3}$
  
  After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is an excerpt from a textbook, "Elementary Linear Algebra 7th Edition," published by Cengage Learning. It contains a math problem from Chapter 1, Section 1.1, Exercise 41, which involves solving a system of linear equations. The excerpt includes the problem statement, the answer, and a step-by-step solution.
  
  Label: question-and-answer