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

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    Algebra Algebra 2
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SUMMARY

The problem involves solving for three numbers, \( x \), \( y \), and \( z \), given the equations \( x + y + z = 95 \), \( y = x + 5 \), and \( z = 3y \). By applying the substitution method, the values are determined as \( x = 15 \), \( y = 20 \), and \( z = 60 \). This systematic approach effectively utilizes algebraic manipulation to arrive at the solution.

PREREQUISITES
  • Understanding of algebraic equations and systems of equations
  • Familiarity with substitution and elimination methods
  • Basic knowledge of mathematical operations (addition, subtraction, multiplication, division)
  • Ability to manipulate variables and expressions
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  • Study the method of solving systems of equations using matrices
  • Learn about the graphical representation of systems of equations
  • Explore advanced algebraic techniques such as the quadratic formula
  • Practice word problems that require the formulation of equations
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Students studying Algebra 2, educators teaching algebraic concepts, and anyone looking to improve their problem-solving skills in mathematics.

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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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