exploring systems of linear equations ( math)

Introduction to Systems of Linear Equations (math)

• Explore systems of linear equations.

• Lesson Objective: By the end, be able to determine if a point is a solution of a system of two linear equations.

• Specific Goals:

  • Determine if a point is a solution of a system of two linear equations.

  • Identify the solution of a system of two linear equations from a graph and verify the solution algebraically.

Vocabulary Words

• Substitute (green): A general academic term used across subjects.

• Intersect (purple): The point where two lines meet in a graph.

• Ordered pair (purple): A pair of numbers (x, y) that represent a point on a Cartesian plane.

• Solution to a system (purple): The ordered pair that satisfies all equations in the system.

• System of linear equations (purple): A set of two or more linear equations with the same variables (e.g., x and y).

Key Question

• How can you determine if a point is a solution to a system of linear equations?

Real-World Example

• Scenario: Buying fruit with a specific amount of money. This represents many real-world situations where systems of equations apply.

Review of Linear Equations

• Solution of a Linear Equation (Example): Solve the equation $3x + y = 6$.

  • For a single variable, solutions are expressed as $x = k$.

  • For two variables: Solutions are ordered pairs ($x, y$) that satisfy the equation.

  • Example Solution: Graph of the equation shows a line of solutions, each satisfying the equation.

Testing a Solution for a Linear Equation

• Testing Point (1, 3) in Equation $3x + y = 6$:

  • Substitute into the equation: $3(1) + 3 = 6$.

  • Calculation: $3 + 3 = 6$.

  • True statement confirmed: (1, 3) is a solution for the equation.

Systems of Linear Equations

• Definition: A system of linear equations consists of two or more linear equations with shared variables (e.g., x, y).

• A solution of a system will also be presented as an ordered pair $(x, y)$, which must satisfy both equations.

Example: Testing a Solution to a System of Equations

• Given equations:

  1. $3x + y = -1$

  2. $y = -x + 1$

• Tested solution point: (-1, 2).

Testing in the First Equation:

• Substitute values: $3(-1) + 2 = -1$.

  • Calculations: $-3 + 2 = -1$

  • True statement: Thus, it satisfies the first equation.

Testing in the Second Equation:

• Substitute values: $2 = -(-1) + 1$.

  • Calculations: $2 = 1 + 1$

  • Confirmed true: therefore, it satisfies the second equation.

• Conclusion: The point (-1, 2) is a solution to the system of equations.

Solutions vs. Non-Solutions

• Example of a solution: (1, 3)

• Example of a non-solution: (-2, -3)

  • Rationale for (1, 3):

  • Substitute into both equations:

    • 1st equation: $x = 1$ gives true statement.

    • 2nd equation: $x = 1$ verifies as well.

• Rationale for (-2, -3):

  • Substitute into both equations:

  • 1st equation: $x = -2$ leads to a true statement.

  • 2nd equation: $x = -2$ yields a false stmt (i.e., not satisfying both equations).

Graphical Representation

• Example of intersection of linear equations:

  • Graphing shows where two lines intersect, representing the solution to the system.

  • Each point on a line is a solution for that equation; hence intersection point is the common solution for both equations.

  • Example: Intersection point (-1, 2) confirmed as the solution.

Verifying Solution from Graph

• Regardless of graph accuracy, represented solutions can be approximated but should be checked algebraically.

Real-World Application Example

• Scenario with Helene organizing a party with fruit:

  • Goal: 10 pieces of fruit representing apples and oranges, with $8 total budget.

  • Equations derived:

  1. $x + y = 10$ (total pieces of fruit)

  2. $x + 0.5y = 8$ (total cost)

• Variables: $x$ = number of apples, $y$ = number of oranges.

• Find the solution by graphing the equations to identify where they intersect.

• Found intersection point: (6, 4).

Understanding Solution Points

• Result: Helene buys 6 apples and 4 oranges.

• Verification:

  • For first equation: $6 + 4 = 10$ (true statement).

  • For second equation: costs yield $6 + 0.5(4) = 8$ (true statement).

• Confirmed: Solution meets both logical and budget constraints.