Solving Systems of Linear Equations by Substitution

Introduction to Systems of Linear Equations via Substitution

• Conceptual Overview: Chapter 9 focuses on solve systems of linear equations using algebraic methods. This section represents the most advanced algebra covered in the course. The primary technique discussed is the method of substitution.

• Definition of Substitution: An algebraic method used to solve systems of equations by expressing one variable in terms of the other and substituting that expression into the second equation to reduce the system to a single-variable equation.

The Five-Step Procedure for Substitution

1. Isolate a Variable: Identify any variable ($x$ or $y$) from either of the two equations and isolate it.

   • Strategic Selection: Always look for the "easiest route" by choosing a variable with no coefficient (an implicit coefficient of $1$). This helps avoid introducing fractions early in the process.

2. Substitute into the Other Equation: Take the expression resulting from step one and substitute it into the other equation in place of that variable. This will yield an equation with only one specific variable.

3. Solve for the First Variable: Use algebraic manipulation to solve for the value of the remaining variable.

4. Solve for the Second Variable: Substitute the numerical value found in step three back into the isolation equation (from step one) to determine the value of the second variable.

5. Verify and Format: Check the resulting values in both original equations to ensure accuracy. Write the final answer as an ordered pair $(x, y)$.

Identifying the Variable to Isolate

• Coefficient Analysis: When examining equations like $x + 4y = 6$, the variable $x$ has an implicit coefficient of $1$.

• Avoiding Fractions: Variables that have explicit coefficients (e.g., $4y$, $2x$, $-3y$) will result in fractions if they are isolated. Therefore, the method of substitution is most efficient when at least one variable in the system lacks an explicit coefficient (meaning the coefficient is $1$ or $-1$).

Example 1: Solving a Basic System

• Given System:

  • Equation 1: $x + 4y = 6$

  • Equation 2: $2x - 3y = 1$

• Step 1: Isolate the variable with no coefficient:

  • Subtract $4y$ from both sides of Equation 1: $x = 6 - 4y$

• Step 2: Substitute into the other equation:

  • Replace the $x$ in Equation 2 with the expression $(6 - 4y)$.

  • Equation: $2(6 - 4y) - 3y = 1$

• Step 3: Solve for $y$:

  • Distribute the $2$: $12 - 8y - 3y = 1$

  • Combine like terms ($y$ terms): $12 - 11y = 1$

  • Move constants to one side: $-11y = 1 - 12$

  • Solve: $-11y = -11 \rightarrow y = 1$

• Step 4: Solve for $x$:

  • Using the isolation formula: $x = 6 - 4(1)$

  • $x = 6 - 4 = 2$

• Final Answer: The solution is the ordered pair $(2, 1)$.

Interpretation of the Solution

• Graphing Connection: The solution to a system of equations represents the point where the two lines intersect on a graph.

• Ordered Pair Format: Solutions must be written as $(x, y)$ where $x$ is always the first value and $y$ is always the second value.

• Verification (The Check):

  • Equation 1: $2 + 4(1) = 6 \rightarrow 6 = 6$ (Correct).

  • Equation 2: $2(2) - 3(1) = 1 \rightarrow 4 - 3 = 1 \rightarrow 1 = 1$ (Correct).

Example 2: Solving with Non-Zero Constants

• Given System:

  • Equation 1 (implied): $7x + y = -2$

  • Equation 2: $5x - 3y - 2 = 0$

• Step 1: Isolate $y$ in Equation 1:

  • $y = -7x - 2$

• Step 2 & 3: Substitution and Solving:

  • Substitute $-7x - 2$ for $y$ in Equation 2: $5x - 3(-7x - 2) - 2 = 0$

  • Distribute $-3$: $5x + 21x + 6 - 2 = 0$

  • Combine like terms: $26x + 4 = 0 \rightarrow 26x = -2$ (Note: Based on transcript narration the constant movement results in $2$ divided by $26$ in the final step).

  • $x = \frac{2}{26}$

  • Lowest Terms: Always reduce fractions. $x = \frac{1}{13}$.

• Step 4: Solve for $y$:

  • $y = -7(\frac{1}{13}) = -\frac{7}{13}$.

• Numerical Standard: Keep answers as fractions in lowest terms. Do not convert to decimals.

• Ordered Pair: $(\frac{1}{13}, -\frac{7}{13})$.

Example 3: Systems with Existing Fractions

• Strategy: Liquidating Denominators: The best method for handling equations with fractions is to multiply the entire equation by the common denominator to create whole-number coefficients.

• Given System:

  • Equation 1: $y = \frac{1}{3}x - 5$

  • Equation 2: $x - \frac{y}{5} = 13$

• Clearing Denominators:

  • Equation 1: Multiply the entire equation by $3$

    • $3y = 1(x) - 15$

    • Result: $3y = x - 15$

  • Equation 2: Multiply the entire equation by $5$

    • $5(x) - 1(y) = 5 \times 13$

    • Result: $5x - y = 65$

• Substitution Process:

  • Isolate $x$ in the simplified Equation 1: $x = 3y + 15$

  • Substitute into Equation 2: $5(3y + 15) - y = 65$

  • Expand: $15y + 75 - y = 65$

  • Combine like terms: $14y + 75 = 65$

  • Solve for $y$: $14y = 65 - 75 \rightarrow 14y = -10$

  • Reduce: $y = -\frac{10}{14} = -\frac{5}{7}$

• Solve for $x$:

  • $x = 3(-\frac{5}{7}) + 15$

  • Using a calculator: $x = \frac{90}{7}$

• Final Solution: $(\frac{90}{7}, -\frac{5}{7})$.