Key rules for solving systems of equations (substitution & elimination)

📚 Understanding Systems of Equations

A system of equations is a set of two or more equations containing the same variables. The 'solution' to a system of equations is the set of values that, when substituted for the variables, makes all the equations true simultaneously. Think of it as finding the sweet spot where all the equations agree! The two most common methods for solving these systems are substitution and elimination.

📜 Historical Context

The concept of solving multiple equations simultaneously dates back to ancient civilizations. Babylonians and Egyptians used techniques to solve problems involving multiple unknowns. However, a more systematic approach developed over time, with significant contributions from mathematicians like Carl Friedrich Gauss, who refined methods for solving linear systems.

🔑 Key Principles: Substitution Method

• 🎯 Isolate a Variable: Choose one equation and solve it for one variable in terms of the other. For example, given $x + y = 5$, you could solve for $x$ to get $x = 5 - y$.
• 🔄 Substitute: Substitute the expression you found in the previous step into the other equation. This will result in a single equation with only one variable.
• 🧩 Solve: Solve the new equation for the remaining variable.
• ↩️ Back-Substitute: Substitute the value you found back into either of the original equations (or the isolated variable equation) to solve for the other variable.
• ✅ Check: Verify your solution by plugging both values into both original equations.

🧮 Key Principles: Elimination Method

• ➕ Align: Arrange the equations so that like terms (e.g., $x$ and $x$, $y$ and $y$) are aligned in columns.
• ⚖️ Multiply (if necessary): Multiply one or both equations by a constant so that the coefficients of one variable are opposites (e.g., $2x$ and $-2x$).
• ➕ Add: Add the equations together. This will eliminate one variable, leaving you with a single equation with one variable.
• 🧩 Solve: Solve the new equation for the remaining variable.
• ↩️ Back-Substitute: Substitute the value you found back into either of the original equations to solve for the other variable.
• ✅ Check: Verify your solution by plugging both values into both original equations.

🌍 Real-World Examples

Example 1: Ticket Sales (Substitution)

Suppose you sell tickets to a school play. Adult tickets cost $5, and student tickets cost $3. If you sell a total of 50 tickets and collect $190, how many of each type of ticket did you sell?

Let $a$ be the number of adult tickets and $s$ be the number of student tickets. The system of equations is:

• $a + s = 50$
• $5a + 3s = 190$

Solving this by substitution: from the first equation, $a = 50 - s$. Substituting into the second equation: $5(50 - s) + 3s = 190$. Simplifying: $250 - 5s + 3s = 190$, which gives $-2s = -60$, so $s = 30$. Then, $a = 50 - 30 = 20$.

Example 2: Mixing Solutions (Elimination)

A chemist needs to prepare 10 liters of a 25% acid solution. She has a 10% acid solution and a 50% acid solution. How many liters of each solution should she mix?

Let $x$ be the liters of 10% solution and $y$ be the liters of 50% solution. The system of equations is:

• $x + y = 10$
• $0.10x + 0.50y = 2.5$ (since 25% of 10 liters is 2.5 liters of acid)

Solving this by elimination: Multiply the first equation by -0.10: $-0.10x - 0.10y = -1$. Add this to the second equation: $0.40y = 1.5$, so $y = 3.75$. Then, $x = 10 - 3.75 = 6.25$.

📝 Practice Quiz

Solve the following systems of equations:

1. Solve:
   • $2x + y = 7$
   • $x - y = -1$
2. Solve:
   • $3x - 2y = 8$
   • $x + 4y = -2$
3. Solve:
   • $x = 2y + 5$
   • $3x - 5y = 16$
4. Solve:
   • $4x + 3y = 11$
   • $2x - y = 1$
5. Solve:
   • $5x - 2y = 1$
   • $2x + 3y = 8$
6. Solve:
   • $y = 3x - 2$
   • $y = -2x + 8$
7. Solve:
   • $x + y = 12$
   • $x - y = 4$

🎯 Conclusion

Mastering substitution and elimination unlocks powerful problem-solving capabilities. Keep practicing, and you'll become a systems-of-equations pro in no time!