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How to Solve Systems of Equations by Elimination (Multiplying One Equation)

Hey everyone! ๐Ÿ‘‹ I'm struggling with solving systems of equations by elimination, especially when I need to multiply one of the equations first. Can anyone break it down in a way that makes sense? ๐Ÿ™
๐Ÿงฎ Mathematics
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๐Ÿ“š Understanding Systems of Equations by Elimination

A system of equations is a set of two or more equations with the same variables. Solving a system means finding values for the variables that make all equations true simultaneously. Elimination is a method to solve these systems by adding or subtracting the equations to eliminate one variable.

๐Ÿ“œ A Brief History

The concept of solving systems of equations dates back to ancient civilizations. Methods for solving linear equations were found in Babylonian tablets and ancient Chinese texts. The development of algebraic notation by mathematicians like Diophantus paved the way for the modern elimination method we use today.

๐Ÿ”‘ Key Principles of Elimination

  • ๐ŸŽฏ Goal: Eliminate one variable by creating opposite coefficients.
  • ๐Ÿ”ข Multiplication: Multiply one or both equations by a constant to make the coefficients of one variable opposites.
  • โž• Addition/Subtraction: Add the equations together. If the coefficients were truly opposites, one variable will be eliminated.
  • โœ… Solve: Solve the resulting equation for the remaining variable.
  • ๐Ÿ”„ Substitution: Substitute the found value back into one of the original equations to solve for the other variable.
  • ๐Ÿง Check: Verify your solution by substituting both values into both original equations.

๐Ÿ“ Step-by-Step Guide: Multiplying One Equation

Hereโ€™s how to solve systems of equations by elimination when you need to multiply just one equation.

  • 1๏ธโƒฃ Inspect: Look at the system and determine which variable is easiest to eliminate. Sometimes you can eliminate a variable without multiplying, but in this case, we need to multiply.
  • โœ–๏ธ Multiply: Choose a factor to multiply one equation so that the coefficients of one variable are opposites. Be sure to distribute the factor to every term in the equation.
  • โž• Add: Add the modified equation to the other original equation. This will eliminate one variable.
  • โœ… Solve: Solve the resulting equation for the remaining variable.
  • โ†ฉ๏ธ Substitute: Substitute the value you found back into either of the original equations to solve for the other variable.
  • ๐Ÿ’ฏ Check: Check your solution by substituting both values into both original equations. If both equations hold true, your solution is correct.

๐Ÿ’ก Example 1:

Solve the system: $2x + y = 7$ $x + 3y = 7$

  1. Multiply the second equation by -2: $-2(x + 3y) = -2(7)$ becomes $-2x - 6y = -14$
  2. Add the modified equation to the first: $(2x + y) + (-2x - 6y) = 7 + (-14)$ $-5y = -7$
  3. Solve for $y$: $y = \frac{-7}{-5} = \frac{7}{5}$
  4. Substitute $y = \frac{7}{5}$ into the first original equation: $2x + \frac{7}{5} = 7$ $2x = 7 - \frac{7}{5} = \frac{35}{5} - \frac{7}{5} = \frac{28}{5}$ $x = \frac{28}{5} \div 2 = \frac{14}{5}$
  5. Solution: $x = \frac{14}{5}, y = \frac{7}{5}$

๐Ÿงฎ Example 2:

Solve the system: $3x - 2y = 10$ $x + y = 5$

  1. Multiply the second equation by 2: $2(x + y) = 2(5)$ becomes $2x + 2y = 10$
  2. Add the modified equation to the first: $(3x - 2y) + (2x + 2y) = 10 + 10$ $5x = 20$
  3. Solve for $x$: $x = \frac{20}{5} = 4$
  4. Substitute $x = 4$ into the second original equation: $4 + y = 5$ $y = 5 - 4 = 1$
  5. Solution: $x = 4, y = 1$

๐ŸŒ Real-World Applications

Systems of equations are used in various fields, including:

  • ๐Ÿงช Chemistry: Balancing chemical equations.
  • ๐Ÿ’ธ Economics: Modeling supply and demand.
  • โœˆ๏ธ Engineering: Designing structures and circuits.
  • ๐Ÿ“ˆ Finance: Portfolio optimization.

โœ๏ธ Practice Quiz

Solve the following systems of equations using the elimination method (multiplying one equation where necessary):

  1. $x + 2y = 7$, $3x + y = 7$
  2. $2x - y = 3$, $x + y = 9$
  3. $4x + 3y = 17$, $x - y = -2$
  4. $5x + 2y = 1$, $2x - y = 3$
  5. $3x + 4y = 6$, $x + 2y = 2$

๐Ÿ”‘ Answer Key

  1. $x = \frac{7}{5}, y = \frac{14}{5}$
  2. $x = 4, y = 5$
  3. $x = 1, y = 3$
  4. $x = 1, y = -2$
  5. $x = 2, y = 0$

๐Ÿ Conclusion

Solving systems of equations by elimination, especially when multiplying one equation, is a fundamental skill in algebra. By understanding the key principles and practicing regularly, you can master this technique and apply it to various real-world problems. Keep practicing, and you'll become an expert in no time!

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