Algebra II Study Notes - Systems of Equations

ALGEBRA II: 6 SYSTEMS OF EQUATIONS: 06.01 SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY

Introduction

  • Scenario: Danielle is playing "Algebra City Defense" where she helps Mathonauts against a giant monster. She must decide between using missiles or ray-guns before furthering her mission.
  • Constraints:
    • Only 7 slots for weapons.
    • 23 batteries available for the operation of the weapons.
  • Essential Questions: After this lesson, students should be able to answer:
    • What methods can be used to solve a system of equations?
    • What types of solutions do these methods yield?

Systems of Equations

  • Definition: A system of equations consists of two or more equations that represent constraints for a situation involving the same variables.
  • Example: For Danielle's game,
    • Missiles: Use 5 batteries.
    • Ray-guns: Use 3 batteries.
  • Equations Set Up: Let m (missiles) and r (ray-guns) represent the variables.
    • Equation 1: m+r=7m + r = 7 (indicating the total slots).
    • Equation 2: 5m+3r=235m + 3r = 23 (indicating battery usage).

Solving Systems of Equations

  • Method Overview: Three standard methods of solving systems of equations are:
    • Elimination
    • Substitution
    • Graphing
  • Personal Preference: Students should identify which method they are most comfortable with.
Example 1: Using Elimination to Solve a System
  • Given Equations:
    1. x+y=9x + y = 9
    2. x−y=3x - y = 3
Step-by-Step Solution:

  1. Identify/Create Opposite Coefficients:

    • Identify opposite coefficients to facilitate elimination. In this example, the coefficients of y are:
    • (+yextand−y)(+y ext{ and } -y)

  2. Add the Equations Vertically:

    • Align like terms:
    • egin{array}{r} x + y = 9 \
      • x - y = 3 \
        \
        ext{ } \
        2x + 0y = 12
        ext{ }
        ext{ }
        ext{ }
        ext{ }
        ext{ }
        ext{ }
        ext{ }

    2x = 12
    ext{ }
    ext{ }
    ext{ }
    ext{ }
    ext{ }

    ext{ }(After adding terms vertically)

    2x = 12

    ext{ }
    ext{ }
    ext{ }

    ext{ }
    ext{ }

    \ \

    ext{ } \

    \ \
    2x = 12
    \ 2x = 12 \

    ext{ }
    ext{ }

    }

  3. Solve for the Variable:

    • Divide by 2;
      x=6x = 6.

  4. Substitute to Solve for the Other Variable:

    • Substitute x back into one of the original equations (e.g., the first equation):
    • 6+y=96 + y = 9
    • y=3y = 3
  • Final Solution: The solution to the system of equations is:
    • x=6,y=3x = 6, y = 3.
Verification:
  • Substitute x and y back into original equations to confirm:
    • First Equation Check:
    • 6+3=9ext(True)6 + 3 = 9 ext{ (True)}
    • Second Equation Check:
    • 6−3=3ext(True)6 - 3 = 3 ext{ (True)}
    • Graphically, the solution is the point (6, 3).

Real-World Applications

The Big Promotion! Example
  • Scenario: Tickets sold at a movie theater: total tickets is 150, adult tickets at $9, child tickets at $6, earning a total of $1,155.
  • Variables Identified:
    • Let a = number of adult tickets sold,
    • Let c = number of child tickets sold.
  • Equations Established:
    • Equation 1: a+c=150a + c = 150 (total tickets sold)
    • Equation 2: 9a+6c=11559a + 6c = 1155 (total revenue from ticket sales)
Steps to Solve:

  1. Create Opposite Coefficients:

    • Multiply the first equation by -9 to eliminate a:
    • −9(a+c=150)-9(a + c = 150)
    • Which transforms to: −9a−9c=−1350-9a - 9c = -1350 (first modified equation)
    • The second equation stays as is:
    • 9a+6c=11559a + 6c = 1155

  2. Vertically Add the Equations:

    • egin{align*}
      (-9a - 9c = -1350) \
      (9a + 6c = 1155) \
      0a - 3c = -195
      ext{ }
      ext{ }
      \
      ext{ }
      0a - 3c = -195 \
      -3c = -195
      ext{ }
      ext{ }
      \

      ext{ }
      c = 65
      ext{ }
      ext{ }
      \
      ext{ }\
      ext{ }\
      ext{ }
      ext{ }
      ext{ }
      \

      ext{ } \ \

    0a - 3c = -195\

    ext{ }\
    \ \

    -3c = -195 \
    c = 65 \
    -9a -9c = -1350 \

    9a + 6c = 1155\

\
-9(a + c = 150)
ext{ }
\ 9a + 6c = 1155 \ \ -9(a + c = 150)\

\ 0a - 3c = -195\ c = 65 \

9a + 6c = 1155 \
0a - 3c = -195\
ext{ } \
\ \

0a - 3c = -195\
\

\ \
ext{ }
c = 65 \
a + c = 150
\ a + 65 = 150\
\
a + 65 = 150\
65 + a= 150\ a + 65 = 150\ a + 65 = 150\
a = 85
ext{ }
ext{ }

a + 65 = 150\
a + 65 = 150\ a = 85 + c = 150\ a = 85\a + 65 = 150\
88 + 65 = 150=
8 5 + c=150\
\ \ \ a = 85
\ \
8 \
\ \ \ a = 65\
\
\ \ \ The final solution is:

  • There were 85 adult tickets sold and 65 child tickets sold.
  • Outcome: Super impressed with your ability, leading to a promotion!

Special Systems of Equations

  • Scenarios of Variables Disappearing:
    • Some systems have the variables vanish, leaving only a statement.
    • Examples Include:
    • When manipulating equations leads to the conclusion of truths such as 0x + 0y = 0 indicative of infinitely many solutions.
    • Alternatively, 0x + 0y = 9 indicating no solution due to contradictory statements.
Infinite Solutions
  • When a system of equations results in a true statement and two equations are equivalent (e.g., representing the same line).
  • Graphically: Both lines overlap producing infinitely many solutions for the system.
No Solution
  • Definitions denote when no values satisfy both equations (e.g., parallel lines with the same slope but differing y-intercepts).
  • Example: For lines y = -x - 1andandy = -x + 2$$; indicate parallelism and no intersection.
Conclusion

Learning how to tackle systems of equations not only enhances mathematical proficiency but also cultivates problem-solving abilities across various disciplines and real-world applications.