Step-by-step guide to solving systems by elimination with multiplication

📚 What is Solving Systems by Elimination with Multiplication?

Solving systems of equations by elimination with multiplication is a method used to find the values of two or more variables that satisfy a set of equations. It involves manipulating one or both equations by multiplying them by a constant so that when the equations are added or subtracted, one variable is eliminated. This allows you to solve for the remaining variable, and then substitute that value back into one of the original equations to solve for the other variable.

📜 A Brief History

The concept of solving systems of equations dates back to ancient civilizations, with early forms of algebraic manipulation found in Babylonian and Egyptian texts. The systematic approach to elimination, however, gained prominence with the development of modern algebra in the 17th and 18th centuries. Mathematicians like Gauss made significant contributions to these techniques, leading to more efficient and generalized methods for solving linear systems.

🔑 Key Principles of Elimination with Multiplication

• 🎯 Goal: To eliminate one variable by creating opposite coefficients.
• 🔢 Multiplication: Multiply one or both equations by a constant to make the coefficients of one variable opposites.
• ➕/➖ 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 back into one of the original equations to find the value of the eliminated variable.

🧑‍🏫 Step-by-Step Guide

1. ✏️ Write the System: Clearly write out the system of equations. For example:
   • Equation 1: $2x + 3y = 13$
   • Equation 2: $5x - 2y = 4$
2. 🎯 Choose a Variable to Eliminate: Decide which variable is easier to eliminate. In this case, let's eliminate $y$.
3. 🔢 Multiply Equations: Multiply Equation 1 by 2 and Equation 2 by 3:
   • Equation 1 (multiplied by 2): $4x + 6y = 26$
   • Equation 2 (multiplied by 3): $15x - 6y = 12$
4. ➕ Add the Equations: Add the modified equations to eliminate $y$:
   • $(4x + 6y) + (15x - 6y) = 26 + 12$
   • $19x = 38$
5. 💡 Solve for $x$: Divide both sides by 19:
   • $x = \frac{38}{19} = 2$
6. 🔄 Substitute $x$ Back: Substitute $x = 2$ into Equation 1:
   • $2(2) + 3y = 13$
   • $4 + 3y = 13$
7. 🔑 Solve for $y$: Solve for $y$:
   • $3y = 9$
   • $y = \frac{9}{3} = 3$
8. ✅ Check Solution: Check the solution in both original equations:
   • Equation 1: $2(2) + 3(3) = 4 + 9 = 13$ (Correct)
   • Equation 2: $5(2) - 2(3) = 10 - 6 = 4$ (Correct)
9. 📝 Final Answer: The solution is $x = 2$ and $y = 3$.

➗ Real-World Examples

• 💰 Finance: Determining the break-even point for investments with different interest rates.
• 🧪 Chemistry: Balancing chemical equations by determining the correct stoichiometric coefficients.
• ✈️ Engineering: Calculating forces in static equilibrium problems, like in bridge design.

✍️ Practice Quiz

1. Solve: $\begin{cases} 3x + 2y = 7 \ 4x - y = -2 \end{cases}$
2. Solve: $\begin{cases} 5x - 3y = 16 \ x + 2y = 11 \end{cases}$
3. Solve: $\begin{cases} 2x + 5y = -4 \ 3x - y = 11 \end{cases}$
4. Solve: $\begin{cases} 4x + 3y = 1 \ 5x + 2y = 4 \end{cases}$
5. Solve: $\begin{cases} 6x - 4y = 10 \ 2x + 3y = 7 \end{cases}$
6. Solve: $\begin{cases} 7x + 2y = 24 \ 5x - 4y = -2 \end{cases}$
7. Solve: $\begin{cases} 8x - 3y = -25 \ 4x + 5y = 1 \end{cases}$

💡 Conclusion

Solving systems of equations by elimination with multiplication is a powerful technique with broad applications. By mastering the steps outlined above, you can confidently tackle a wide range of problems in mathematics, science, and engineering. Keep practicing, and you'll become a pro in no time!