The Substitution Method - Algebra-1

1. Fundamental Concepts

Core Definition: The Substitution Method is an algebraic approach to solving systems of linear equations in two variables. Its core idea is to express one variable in terms of the other using one of the equations, then substitute this expression into the other equation. This eliminates one variable, converting the system into a single-variable linear equation that can be solved.
Applicable Scenarios: It is particularly convenient when at least one equation in the system can be easily transformed into a form where one variable is expressed in terms of the other (e.g., \(y = ax + b\) or \(x = ay + b\)).

Elimination Goal: Transform the system of linear equations in two variables into a linear equation in one variable through substitution, reducing the number of unknowns for step-by-step solving.
Core Steps:
1. Select one equation from the system and rearrange it to express one variable (e.g., y) in terms of the other variable (e.g., x) (i.e., "express the variable").
2. Substitute this expression into the other equation, replacing the corresponding variable, resulting in a linear equation with only one unknown (i.e., "substitute and eliminate").
3. Solve this one-variable linear equation to find the value of one variable.
4. Substitute the found value back into the previously derived expression to find the value of the other variable.
5. Verify that the solution satisfies both equations of the original system (an optional step but helpful to ensure accuracy).

System of Equations:\(\begin{cases}y = 3x \\ x + y = 8\end{cases}\)

Solution:

The first equation is already in the form \(y = 3x\).
Substitute \(y = 3x\) into the second equation: \(x + 3x = 8\).
Simplify to get \(4x = 8\), so \(x = 2\).
Substitute \(x = 2\) into \(y = 3x\), we get \(y = 6\).
The solution is \(\begin{cases}x = 2 \\ y = 6\end{cases}\).

System of Equations:\(\begin{cases}2x + y = 5 \\ y = -3x + 6\end{cases}\)

Solution:

The second equation is already in the form \(y = -3x + 6\).
Substitute \(y = -3x + 6\) into the first equation: \(2x + (-3x + 6) = 5\).
Simplify to get \(-x + 6 = 5\), so \(x = 1\).
Substitute \(x = 1\) into \(y = -3x + 6\), we get \(y = 3\).
The solution is \(\begin{cases}x = 1 \\ y = 3\end{cases}\).

System of Equations:\(\begin{cases}3x - 2y = 1 \\ x = \frac{y + 4}{2}\end{cases}\)

Solution:

The second equation is already in the form \(x = \frac{y + 4}{2}\).
Substitute \(x = \frac{y + 4}{2}\) into the first equation: \(3\times\frac{y + 4}{2} - 2y = 1\).
Multiply both sides by 2 to eliminate the denominator: \(3(y + 4) - 4y = 2\).
Expand and simplify: \(3y + 12 - 4y = 2\), which simplifies to \(-y + 12 = 2\), so \(y = 10\).
Substitute \(y = 10\) into \(x = \frac{y + 4}{2}\), we get \(x = 7\).
The solution is \(\begin{cases}x = 7 \\ y = 10\end{cases}\).

Prioritize Simple Expressions: If one equation in the system is in the form of \(y = ax + b\) or \(x = ay + b\), prioritize using this equation for substitution to reduce calculations.
Avoid Complex Fractions: If fractions appear after substitution, eliminate them by multiplying both sides by the least common multiple of the denominators to simplify calculations.
Verify the Solution: After finding the solution, always substitute it back into both equations of the original system to check if both sides are equal (especially useful for complex calculations).
Handle Special Cases: If substitution results in an identity (e.g., \(0 = 0\)), the system has infinitely many solutions. If it results in a contradiction (e.g., \(2 = 3\)), the system has no solution.