Help? Algebra 2 Math - Solve X,Y,Z

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    Algebra Algebra 2
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Discussion Overview

The discussion revolves around solving a word problem involving three numbers whose relationships are defined by a set of equations. The problem is framed within the context of Algebra 2, focusing on the formulation and solution of a system of equations.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents the problem as a system of equations: \( x + y + z = 95 \), \( y = x + 5 \), and \( z = 3y \).
  • The same participant outlines a substitution method to solve the equations, leading to the values \( y = 20 \), \( x = 15 \), and \( z = 60 \).
  • Another participant echoes the initial approach and solution steps, reiterating the same equations and results without introducing new information.
  • A later comment introduces a humorous note about the nature of asking for help, suggesting that repeated assistance can lead to dependency.

Areas of Agreement / Disagreement

Participants generally agree on the formulation of the problem and the solution process, but there is no explicit consensus on the appropriateness of the assistance provided, as indicated by the humorous remarks about "spoon feeding."

Contextual Notes

The discussion does not address potential limitations or assumptions in the problem-solving approach, nor does it explore alternative methods or solutions.

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The sum of three numbers is 95. The second number is 5 more than the first. The third number is 3 times the second. What are the numbers?
 
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Begin by turning the word problem into a system of equations:

Let \( x + y + z = 95 \), \( y = x + 5 \), and \( z = 3y \).

You can now use Elimination, Substitution, or Matrices to solve. I will use substitution by taking the 2nd and 3rd equations and getting the 1st equation in terms of \( y \).

\( y = x+ 5 \) subtract \( 5 \) from both sides.
\( x = y - 5 \)

Substitute \( x = y - 5 \) and \( z = 3y \) into \( x + y + z = 95 \):

\( ( y -5 ) + y + (3y) = 95 \) combine like terms
\( 5y - 5 = 95 \) add \( 5 \) to both sides
\( 5y = 100 \) divide by \( 5 \) on both sides
\( y = 20 \)

Substitute \( y = 20 \) into \( x = y - 5 \):
\( x = (20) - 5 \) simplify
\( x = 15 \)

Substitute \( y = 20 \) into \( z = 3y \):
\( z = 3(20) \) simplify
\( z = 60 \)

ANSWER: \( x = 15 \), \( y = 20 \), and \( z = 60 \)
 
Beer soaked comment follows.
SquareOne said:
Begin by turning the word problem into a system of equations:

Let \( x + y + z = 95 \), \( y = x + 5 \), and \( z = 3y \).

You can now use Elimination, Substitution, or Matrices to solve. I will use substitution by taking the 2nd and 3rd equations and getting the 1st equation in terms of \( y \).

\( y = x+ 5 \) subtract \( 5 \) from both sides.
\( x = y - 5 \)

Substitute \( x = y - 5 \) and \( z = 3y \) into \( x + y + z = 95 \):

\( ( y -5 ) + y + (3y) = 95 \) combine like terms
\( 5y - 5 = 95 \) add \( 5 \) to both sides
\( 5y = 100 \) divide by \( 5 \) on both sides
\( y = 20 \)

Substitute \( y = 20 \) into \( x = y - 5 \):
\( x = (20) - 5 \) simplify
\( x = 15 \)

Substitute \( y = 20 \) into \( z = 3y \):
\( z = 3(20) \) simplify
\( z = 60 \)

ANSWER: \( x = 15 \), \( y = 20 \), and \( z = 60 \)
Prepare thyself for more questions.
Spoon feeding can be very addictive.
 
jonah said:
Prepare thyself for more questions.
Spoon feeding can be very addictive.
Spoon feeding can be very "additive." (Dance)

-Dan
 

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