๐ Definition of the Substitution Method
The substitution method is a technique used to solve systems of linear equations. A system of equations is a set of two or more equations with the same variables. The goal is to find values for the variables that satisfy all equations simultaneously. The substitution method works by solving one equation for one variable and then substituting that expression into the other equation. This results in a single equation with a single variable, which can then be solved. Finally, the value of that variable is substituted back into either of the original equations to find the value of the other variable.
๐ History and Background
The concept of solving simultaneous equations has been around for centuries. Ancient Babylonians and Greeks solved similar problems, though not with the same algebraic notation we use today. The formalization of the substitution method as a distinct algebraic technique developed alongside the growth of symbolic algebra in the 16th and 17th centuries. Mathematicians like Vieta and Descartes contributed to the development of algebraic notation and methods for solving equations, laying the groundwork for the systematic approach we now know as the substitution method.
๐ Key Principles of the Substitution Method
- ๐ Isolate a Variable: Choose one equation and solve it for one of its variables. This means getting one variable alone on one side of the equation. For example, in the equation $x + y = 5$, you could solve for $x$ to get $x = 5 - y$.
- ๐ Substitute: Substitute the expression you found in the previous step into the other equation. If you solved for $x$ in the first equation, substitute that expression for $x$ in the second equation. This will give you a new equation with only one variable.
- ๐งฎ Solve: Solve the new equation for the remaining variable. This will give you a numerical value for that variable.
- โฉ๏ธ Back-Substitute: Substitute the value you found back into either of the original equations (or the rearranged equation from the first step) to solve for the other variable.
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Check: Check your solution by plugging both values into both original equations to make sure they are both satisfied.
๐ Real-World Examples
Example 1: Buying Fruits
Suppose you go to a fruit stand. You buy 2 apples and 3 bananas for $5. Later, you buy 1 apple and 2 bananas for $3. How much does each apple and each banana cost?
Let $a$ be the cost of an apple and $b$ be the cost of a banana. We can set up the following system of equations:
- ๐ $2a + 3b = 5$
- ๐ $a + 2b = 3$
Solve the second equation for $a$: $a = 3 - 2b$.
Substitute into the first equation: $2(3 - 2b) + 3b = 5$.
Simplify and solve for $b$: $6 - 4b + 3b = 5 \Rightarrow -b = -1 \Rightarrow b = 1$.
Substitute $b = 1$ back into $a = 3 - 2b$: $a = 3 - 2(1) = 1$.
So, an apple costs $1 and a banana costs $1.
Example 2: Mixing Solutions
A chemist needs to create 100 mL of a 30% acid solution. They have a 20% acid solution and a 50% acid solution. How much of each solution should they mix?
Let $x$ be the amount of the 20% solution and $y$ be the amount of the 50% solution.
- ๐งช $x + y = 100$
- โ๏ธ $0.20x + 0.50y = 0.30(100)$
Solve the first equation for $x$: $x = 100 - y$.
Substitute into the second equation: $0.20(100 - y) + 0.50y = 30$.
Simplify and solve for $y$: $20 - 0.20y + 0.50y = 30 \Rightarrow 0.30y = 10 \Rightarrow y = \frac{100}{3} \approx 33.33$ mL.
Substitute $y \approx 33.33$ back into $x = 100 - y$: $x \approx 100 - 33.33 = 66.67$ mL.
So, the chemist should mix approximately 66.67 mL of the 20% solution and 33.33 mL of the 50% solution.
๐ก Conclusion
The substitution method is a powerful tool for solving systems of linear equations. By isolating a variable and substituting, you can reduce a system of two equations to a single equation, making it easier to solve. Understanding the principles and practicing with real-world examples can help you master this essential algebraic technique.