6.4 Systems of Linear Equations: Graphing, Substitution, & Elimination Methods

6.4 Systems of Linear Equations: Graphing, Substitution, & Elimination Methods

Many practical scenarios involve multiple interacting conditions with two unknown quantities. In algebra, a system of linear equations (also called simultaneous equations) consists of two or more linear equations containing the same set of variables:

$$\begin{cases} A_1 x + B_1 y = C_1 \\ A_2 x + B_2 y = C_2 \end{cases}$$

A solution to a system is an ordered pair $(x, y)$ that makes both equations true simultaneously. On the TABE 13&14 assessment, questions test your ability to solve systems using three distinct methods—graphing, substitution, and elimination—and model real-world trade and business situations.

---

1. Geometric Interpretation & System Classifications

When two linear equations in two variables are plotted on the Cartesian coordinate plane, their geometric relationship determines the number of solutions:

Three Geometric Outcomes for Systems of Linear Equations:

1. Intersecting Lines          2. Parallel Lines             3. Coincident Lines
        │                             │                             │
      ──┼──                         ═════                         ═════ (Line 1 & 2
       ╱│╲                          ─────                         ─────  overlap exactly)
        │                             │                             │
   One Solution (x,y)            No Solution (∅)             Infinite Solutions (ℝ)
(Consistent & Independent)       (Inconsistent)            (Consistent & Dependent)

---

2. Method 1: The Graphing Method

To solve a system by graphing:

1. Convert both equations to slope-intercept form ($y = mx + b$).
2. Plot the $y$-intercept $(0, b)$ and use the slope $m = \frac{\text{rise}}{\text{run}}$ to locate additional points.
3. Draw straight lines through each set of points.
4. Identify the point of intersection $(x, y)$ and verify algebraically in both equations.

$$\begin{cases} y = 2x - 1 \\ y = -x + 5 \end{cases}$$

• Equating both expressions for $y$: $2x - 1 = -x + 5 \implies 3x = 6 \implies x = 2$.
• Substitute $x = 2$: $y = 2(2) - 1 = 3$. The lines intersect at the point $(2, 3)$.

3. Method 2: The Substitution Method

Substitution is the most direct algebraic method when one variable has a coefficient of $1$ or $-1$ (e.g., $x = \dots$ or $y = \dots$).

The 4-Step Substitution Algorithm

1. Isolate: Solve one of the equations for either variable ($x$ or $y$).
2. Substitute: Replace that variable in the other equation with the isolated expression.
3. Solve: Solve the resulting single-variable equation.
4. Back-Substitute: Plug the obtained value back into the Step 1 expression to find the second variable.

Worked Example: Substitution Method

Problem: Solve the system:

$$\begin{cases} x + 3y = 7 \quad \text{--- (Equation 1)} \\ 2x - 5y = -19 \quad \text{--- (Equation 2)} \end{cases}$$

• Step 1 (Isolate $x$ in Eq 1): $$x = 7 - 3y$$
• Step 2 (Substitute into Eq 2): $$2(7 - 3y) - 5y = -19$$
• Step 3 (Solve for $y$): $$\begin{aligned} 14 - 6y - 5y &= -19 \\ 14 - 11y &= -19 \\ -11y &= -19 - 14 \\ -11y &= -33 \\ y &= 3 \end{aligned}$$
• Step 4 (Back-substitute $y = 3$ to find $x$): $$x = 7 - 3(3) = 7 - 9 = -2$$
• Final Solution: $(-2, 3)$.
• Check: In Eq 2: $2(-2) - 5(3) = -4 - 15 = -19$ (True).

---

4. Method 3: The Elimination (Addition) Method

Elimination is ideal when equations are written in standard form ($Ax + By = C$) and coefficients are integers.

The 4-Step Elimination Algorithm

1. Align: Ensure both equations are written in standard form $Ax + By = C$.
2. Scale: Multiply one or both equations by non-zero constants so that the coefficients of one variable are additive opposites (e.g., $+6y$ and $-6y$).
3. Add & Eliminate: Add the two equations vertically to eliminate that variable.
4. Solve & Back-Substitute: Solve for the remaining variable, then substitute back to find the second coordinate.

Worked Example: Elimination Method

Problem: Solve the system:

$$\begin{cases} 3x + 4y = 18 \quad \text{--- (Eq 1)} \\ 5x - 2y = 4 \quad \text{--- (Eq 2)} \end{cases}$$

• Step 1 & 2 (Multiply Eq 2 by $2$ to align $y$-coefficients): $$\begin{aligned} 3x + 4y &= 18 \\ 2(5x - 2y &= 4) \implies 10x - 4y = 8 \end{aligned}$$
• Step 3 (Add the equations vertically): $$\begin{array}{rrcr} & 3x + 4y & = & 18 \\ + & 10x - 4y & = & 8 \\ \hline & 13x + 0 & = & 26 \\ & 13x & = & 26 \\ & x & = & 2 \end{array}$$
• Step 4 (Back-substitute $x = 2$ into Eq 1): $$3(2) + 4y = 18 \implies 6 + 4y = 18 \implies 4y = 12 \implies y = 3$$
• Final Solution: $(2, 3)$.

---

5. Identifying Special Systems Algebraically

• Inconsistent System (No Solution): If variable terms eliminate and produce a false contradiction ($0 = 8$), the lines are parallel and have no solution.

  $$\begin{cases} 2x - y = 4 \\ 4x - 2y = 14 \end{cases} \implies -2(2x - y = 4) \implies -4x + 2y = -8$$ $$\text{Adding to Eq 2:} \quad (-4x + 4x) + (2y - 2y) = -8 + 14 \implies 0 = 6 \quad (\text{No Solution})$$

• Dependent System (Infinitely Many Solutions): If variable terms eliminate and produce a true tautology ($0 = 0$), the equations represent identical lines.

  $$\begin{cases} x - 2y = 3 \\ 3x - 6y = 9 \end{cases} \implies -3(x - 2y = 3) \implies -3x + 6y = -9$$ $$\text{Adding to Eq 2:} \quad (-3x + 3x) + (6y - 6y) = -9 + 9 \implies 0 = 0 \quad (\text{Infinitely Many Solutions})$$

---

6. Real-World Applications & Word Problems

Application 1: Ticket Sales / Total Value Problems

Problem: A performing arts center sold a total of $450$ tickets to a weekend concert, collecting $\text{\textdollar}3,700$ in total box office revenue. Student tickets cost $\text{\textdollar}6$ each, and general admission adult tickets cost $\text{\textdollar}10$ each. How many of each ticket type were sold?

1. Define Variables: Let $s = \text{student tickets}$, $a = \text{adult tickets}$.
2. Formulate System: $$\begin{cases} s + a = 450 \quad (\text{Total Ticket Count}) \\ 6s + 10a = 3700 \quad (\text{Total Dollar Revenue}) \end{cases}$$
3. Solve via Substitution: From Eq 1, $s = 450 - a$. $$\begin{aligned} 6(450 - a) + 10a &= 3700 \\ 2700 - 6a + 10a &= 3700 \\ 2700 + 4a &= 3700 \\ 4a &= 1000 \\ a &= 250 \text{ adult tickets} \end{aligned}$$
4. Find $s$: $s = 450 - 250 = 200\text{ student tickets}$.

Application 2: Mixture Problems

Problem: A medical clinic needs $60\text{ liters}$ of a $25\%$ antiseptic solution. The clinic has a $10\%$ stock solution and a $40\%$ stock solution on hand. How many liters of each must be mixed?

1. Define Variables: Let $x = \text{liters of 10\% solution}$, $y = \text{liters of 40\% solution}$.
2. Set Up System: $$\begin{cases} x + y = 60 \quad (\text{Total Liquid Volume}) \\ 0.10x + 0.40y = 0.25(60) = 15 \quad (\text{Pure Antiseptic Content}) \end{cases}$$
3. Solve: Multiply concentration equation by $10$: $x + 4y = 150$. Subtract $x + y = 60$: $$(x - x) + (4y - y) = 150 - 60 \implies 3y = 90 \implies y = 30\text{ liters}$$
4. Find $x$: $x = 60 - 30 = 30\text{ liters}$. Conclusion: Combine $30\text{ liters}$ of the $10\%$ solution with $30\text{ liters}$ of the $40\%$ solution.