๐ Understanding Systems of Linear Equations
A system of linear equations is a set of two or more linear equations that are considered together. The solution to a system of linear equations is the set of values that satisfy all equations simultaneously. These systems are incredibly useful for modeling and solving real-world problems involving multiple variables and constraints.
๐ History and Background
The concept of solving systems of equations dates back to ancient civilizations. Egyptians and Babylonians used methods to solve linear equations, although not in the modern algebraic notation we use today. The development of algebraic notation and methods like Gaussian elimination in later centuries provided more systematic ways to solve these systems.
๐ Key Principles
- โ๏ธ Definition: A system of linear equations consists of two or more linear equations with the same variables.
- ๐ฏ Solution: A solution to the system is a set of values for the variables that satisfies all equations.
- ๐ Graphical Interpretation: Each linear equation represents a line. The solution to a system of two linear equations in two variables is the point where the lines intersect.
- โ Substitution Method: Solve one equation for one variable and substitute that expression into the other equation.
- โ Elimination Method: Add or subtract multiples of the equations to eliminate one variable.
- ๐ Matrix Representation: Represent the system as a matrix equation and use matrix operations to solve.
๐ Real-World Examples
Example 1: Mixing Solutions
A chemist needs to prepare 100 ml of a 30% acid solution. They have a 10% acid solution and a 50% acid solution. How many ml of each solution should they mix?
Let $x$ be the amount of 10% solution and $y$ be the amount of 50% solution.
We have the following system of equations:
- $x + y = 100$ (Total volume)
- $0.10x + 0.50y = 0.30(100)$ (Acid concentration)
Solving this system:
From the first equation, $x = 100 - y$. Substituting into the second equation:
$0.10(100 - y) + 0.50y = 30$
$10 - 0.10y + 0.50y = 30$
$0.40y = 20$
$y = 50$
So, $x = 100 - 50 = 50$
Therefore, the chemist should mix 50 ml of the 10% solution and 50 ml of the 50% solution.
Example 2: Cost and Revenue
A business sells two types of products, A and B. The cost to produce each unit of product A is $5, and the cost to produce each unit of product B is $8. The company has a total budget of $1500 for production. They want to produce a total of 200 units. How many units of each product should they produce?
Let $x$ be the number of units of product A and $y$ be the number of units of product B.
We have the following system of equations:
- $x + y = 200$ (Total units)
- $5x + 8y = 1500$ (Total cost)
Solving this system:
From the first equation, $x = 200 - y$. Substituting into the second equation:
$5(200 - y) + 8y = 1500$
$1000 - 5y + 8y = 1500$
$3y = 500$
$y = \frac{500}{3} \approx 166.67$
Since the number of units must be an integer, we can round to $y = 167$
So, $x = 200 - 167 = 33$
Therefore, the business should produce approximately 33 units of product A and 167 units of product B.
Example 3: Distance, Rate, and Time
Two cars start from the same point and travel in opposite directions. One car travels at 60 mph, and the other travels at 40 mph. How long will it take for them to be 300 miles apart?
Let $t$ be the time in hours.
The distance traveled by the first car is $60t$, and the distance traveled by the second car is $40t$.
We have the equation:
$60t + 40t = 300$
$100t = 300$
$t = 3$
Therefore, it will take 3 hours for the cars to be 300 miles apart.
๐ก Conclusion
Systems of linear equations are powerful tools for modeling and solving real-world problems. By understanding the key principles and practicing with examples, you can apply these techniques to a wide range of scenarios. Whether it's mixing solutions, managing costs, or calculating distances, systems of linear equations provide a structured approach to finding solutions.