Classifying Solutions to Systems of Equations

Learning Outcomes

• Learn various methods for solving systems of equations to identify inconsistent systems.

• Inconsistent Systems: Consist of parallel lines with the same slope but different y-intercepts. These lines never intersect.

  • Example: A system like $12 = 0$ indicates no solution.

Key Learning Outcomes

• Determine if a system has:

  • One solution

  • Many solutions (dependent)

  • No solution (inconsistent)

• Express the general solution of a dependent system.

• Interpret solutions concerning business profits.

• Model real situations using systems of equations.

Example: Solving an Inconsistent System of Equations

• Given System:

  • $x = 9 - 2y$

  • $x + 2y = 13$

• Method: Use substitution (as one equation solves for x).

Step-by-Step Solution:

1. Substitute for $x$ in the second equation:

   • Given:
     $x + 2y = 13$
     Substitute $x$:
     $9 - 2y + 2y = 13$

2. Simplify:

   • $9 = 13$ (a false statement, indicating no solution).

Alternative Approach: Slope-Intercept Form

• Convert both equations:

1. First Equation: $x = 9 - 2y$

   - Rearranged: $2y = -1x + 9$

   
   Final Slope-Intercept Form:

   $y = -\frac{1}{2}x + \frac{9}{2}$

2. Second Equation: $x + 2y = 13$
   - Rearranged: $2y = -x + 13$

   
   Final Slope-Intercept Form:
   $y = -\frac{1}{2}x + \frac{13}{2}$ ​

• Comparison: Same slope $(-\frac{1}{2})$, different intercepts → Confirmed parallel lines, inconsistent system.

Dependent Systems of Equations

• A dependent system has an infinite number of solutions (the same line).

• Resulting equation is an identity (e.g., $0 = 0$).

• General solution form:

  • $(x, -\frac{1}{3}x + \frac{2}{3})$

Example: Profit Model for a Business

• Revenue Function: $R = xp$ where $x$ = quantity, $p$ = price.

• Cost Function: Include fixed and variable costs.

• Break-even Point: Where cost = revenue.

  • Example in a graph: If selling 700 units leads to equal cost and revenue, no profit or loss.

• Profit Function: $P(x) = R(x) - C(x)$, crucial for businesses.

Finding Break-Even Point and Profit Function Using Substitution

1. Cost Function: $C(x) = 0.85x + 35000$

2. Revenue Function: $R(x) = 1.55x$

3. Set them equal for break-even:

   • $0.85x + 35000 = 1.55x$

   • Solve for $x$ yields 50000 units, profit = $0.7x - 35000$.

Writing Systems of Linear Equations from Given Situations

1. Identify the total attendance, revenue, cost, etc.

2. Use algebra to set up equations based on that information.

3. Analyze and solve through substitution or elimination.

• Example Circus Revenue:

  • Children’s ticket: $25$, adult ticket: $50$.

  • $c + a = 2000$; $25c + 50a = 70000$

  • Solve yields 1200 children, 800 adults.

Practice Problems

1. For meal tickets: Total $1650$ tickets totaling $14200$; find child/adult tickets sold.

2. For a truck rental decision problem, set costs from two companies as functions and find where one is cheaper than the other.

Chemical Mixture Problem

• Mixing solutions: 70ml of 50% methane. Add x ml of 80% to get 60% final solution.

  1. Setup equation based on amounts and percentages:

     • $0.5 * 70 + 0.8 * x = 0.6 * (70 + x)$

  2. Solve for $x$ yields 35ml required.

• This can also help calculate costs or pricing in business scenarios.

Keep practicing the translation of real situations into mathematical models for better understanding!