Definition of solving systems by elimination using multiplication

📚 Definition of Solving Systems by Elimination Using Multiplication

Solving systems of equations by elimination using multiplication is a method used to find the values of variables (usually $x$ and $y$) that satisfy two or more equations simultaneously. The core idea is to manipulate one or both equations by multiplying them by a constant, such that when you add or subtract the equations, one of the variables cancels out. This allows you to solve for the remaining variable, and then substitute that value back into one of the original equations to find the value of the other variable.

📜 History and Background

The concept of solving systems of equations dates back to ancient civilizations, with early forms found in Babylonian and Chinese mathematics. The systematic approach of elimination evolved over centuries, becoming a fundamental tool in algebra. The use of multiplication to prepare equations for elimination is a refinement that enhances the method's applicability to a wider range of systems.

🔑 Key Principles

• 🎯 Goal: To eliminate one variable by making its coefficients opposites or identical in both equations.
• 🔢 Multiplication: Multiply one or both equations by a carefully chosen constant so that the coefficients of one variable are additive inverses (e.g., 2 and -2) or the same.
• ➕/➖ Addition/Subtraction: Add or subtract the modified equations to eliminate one variable.
• 💡 Solve: Solve the resulting equation for the remaining variable.
• 🔄 Substitution: Substitute the value obtained back into one of the original equations to solve for the other variable.
• ✅ Check: Verify your solution by substituting both values into both original equations.

🧮 Example 1: A Simple Case

Solve the following system of equations:

Equation 1: $x + 2y = 7$

Equation 2: $3x - y = -3$

1. Multiply Equation 2 by 2: $2 * (3x - y) = 2 * (-3)$ becomes $6x - 2y = -6$
2. Add the modified Equation 2 to Equation 1: $(x + 2y) + (6x - 2y) = 7 + (-6)$ simplifies to $7x = 1$
3. Solve for $x$: $x = \frac{1}{7}$
4. Substitute $x = \frac{1}{7}$ into Equation 1: $\frac{1}{7} + 2y = 7$
5. Solve for $y$: $2y = 7 - \frac{1}{7} = \frac{48}{7}$, so $y = \frac{24}{7}$
6. Solution: $x = \frac{1}{7}$, $y = \frac{24}{7}$

➗ Example 2: When Both Equations Need Multiplication

Solve the following system of equations:

Equation 1: $2x + 3y = 8$

Equation 2: $5x - 2y = 1$

1. Multiply Equation 1 by 2: $2 * (2x + 3y) = 2 * 8$ becomes $4x + 6y = 16$
2. Multiply Equation 2 by 3: $3 * (5x - 2y) = 3 * 1$ becomes $15x - 6y = 3$
3. Add the modified equations: $(4x + 6y) + (15x - 6y) = 16 + 3$ simplifies to $19x = 19$
4. Solve for $x$: $x = 1$
5. Substitute $x = 1$ into Equation 1: $2 * 1 + 3y = 8$
6. Solve for $y$: $3y = 8 - 2 = 6$, so $y = 2$
7. Solution: $x = 1$, $y = 2$

🌍 Real-World Applications

• 💰 Finance: Solving for investment portfolios to determine asset allocation.
• 🧪 Chemistry: Balancing chemical equations.
• 🚀 Engineering: Analyzing forces in structural systems.
• 📊 Economics: Modeling supply and demand curves.

📝 Conclusion

Solving systems of equations by elimination using multiplication is a powerful algebraic technique. By strategically manipulating equations, we can efficiently find solutions that satisfy multiple conditions, making it invaluable in various mathematical and real-world scenarios. Mastering this method equips you with a versatile problem-solving skill applicable across numerous disciplines.