3.5 Systems of Linear & Nonlinear Equations and Inequalities

3.5 Systems of Linear & Nonlinear Equations and Inequalities

A system of equations consists of two or more equations sharing common variables. A solution to a system is an ordered pair $(x, y)$ or ordered triple $(x, y, z)$ that satisfies all equations in the system simultaneously.

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1. $2 \times 2$ Linear Systems & Consistency Classification

A $2 \times 2$ system of linear equations features two equations in two variables:

$\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}$

Geometric Classification of Linear Systems

Primary Algebraic Solution Methods

• Substitution: Isolate one variable in an equation and substitute its expression into the second equation.
• Elimination (Addition): Multiply equations by non-zero constants so that the coefficients of one variable are additive opposites, then add the equations to eliminate that variable.

Worked Example 1: Elimination Method

Solve the system: $\begin{cases} 3x + 2y = 12 \quad (1) \\ 5x - 4y = 9 \quad (2) \end{cases}$

1. Multiply equation (1) by $2$: $6x + 4y = 24$.
2. Add to equation (2): $(6x + 4y) + (5x - 4y) = 24 + 9 \implies 11x = 33 \implies x = 3$
3. Substitute $x = 3$ into equation (1): $3(3) + 2y = 12 \implies 9 + 2y = 12 \implies 2y = 3 \implies y = \frac{3}{2}$ Solution: $\left(3, \frac{3}{2}\right)$. System is consistent and independent.

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2. $3 \times 3$ Linear Systems & Sequential Elimination

A system of three linear equations in three variables ($x, y, z$) represents three planes in three-dimensional space.

Sequential Elimination Algorithm ($3 \times 3 \to 2 \times 2 \to 1 \times 1$)

1. Select Target Variable: Choose a variable to eliminate across two pairs of equations.
2. Reduce to $2 \times 2$: Pair Eq 1 and Eq 2 to eliminate the target variable. Pair Eq 1 (or 2) and Eq 3 to eliminate the same variable, forming a $2 \times 2$ system.
3. Solve $2 \times 2$ System: Isolate the two remaining variables.
4. Back-Substitute: Substitute known variables back into an original equation to find the third variable.

Worked Example 2: $3 \times 3$ System

Problem: Solve the system: $\begin{cases} x + y + z = 6 \quad (1) \\ 2x - y + z = 3 \quad (2) \\ x + 2y - z = 2 \quad (3) \end{cases}$

Step-by-Step Solution:

1. Eliminate $z$ using Eq 1 and Eq 3: Add Eq 1 and Eq 3: $(x + y + z) + (x + 2y - z) = 6 + 2 \implies 2x + 3y = 8 \quad (4)$
2. Eliminate $z$ using Eq 2 and Eq 3: Add Eq 2 and Eq 3: $(2x - y + z) + (x + 2y - z) = 3 + 2 \implies 3x + y = 5 \quad (5)$
3. Solve the $2 \times 2$ system (Eq 4 and Eq 5): Multiply Eq 5 by $-3$: $-9x - 3y = -15$. Add to Eq 4: $(2x + 3y) + (-9x - 3y) = 8 - 15 \implies -7x = -7 \implies x = 1$.
4. Substitute $x = 1$ into Eq 5: $3(1) + y = 5 \implies y = 2$.
5. Back-substitute $x = 1, y = 2$ into Eq 1: $1 + 2 + z = 6 \implies z = 3$. Solution: $(1, 2, 3)$.

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3. Systems of Nonlinear Equations

A nonlinear system contains at least one non-linear equation (such as a parabola $y = ax^2 + bx + c$ or circle $x^2 + y^2 = r^2$). A line and parabola can intersect at 0, 1, or 2 points.

Worked Example 3: Line and Parabola System

Problem: Solve the system: $\begin{cases} y = 2x + 1 \quad (1) \\ y = x^2 - 2x + 4 \quad (2) \end{cases}$

Step-by-Step Solution:

1. Equate the expressions for $y$: $x^2 - 2x + 4 = 2x + 1$
2. Rearrange into standard quadratic form: $x^2 - 4x + 3 = 0$
3. Factor the quadratic: $(x - 1)(x - 3) = 0 \implies x = 1 \quad \text{or} \quad x = 3$
4. Find corresponding $y$-values using $y = 2x + 1$:
   • For $x = 1$: $y = 2(1) + 1 = 3 \implies (1, 3)$
   • For $x = 3$: $y = 2(3) + 1 = 7 \implies (3, 7)$ Solution set: $\{(1, 3), (3, 7)\}$.

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4. Systems of Linear Inequalities & Feasible Regions

A system of linear inequalities defines a bounded or unbounded region of the $xy$-plane called the feasible region.

Graphing Rules

1. Boundary Lines: Draw boundary lines as solid for $\le, \ge$ (points included) and dashed for $<, >$ (points excluded).
2. Half-Plane Shading: Test an origin point $(0,0)$ to determine which side of each line to shade.
3. Feasible Region: The solution set is the overlapping region where all shaded half-planes intersect.
4. Corner Vertices: Find intersection points of boundary lines by solving their linear equations as systems.

Worked Example 4: Graphing a System of Linear Inequalities

Problem: Describe the feasible region of the system, and find its corner vertices. $\begin{cases} 2x + y \le 8 \\ x - y < 2 \\ x \ge 0 \end{cases}$

Step-by-Step Solution:

1. Convert each boundary to slope-intercept form and choose the line style:
   • $2x + y = 8 \implies y = -2x + 8$. The operator is $\le$, so the line is solid.
   • $x - y = 2 \implies y = x - 2$. The operator is strict $<$, so the line is dashed.
   • $x = 0$ is the vertical $y$-axis. The operator is $\ge$, so it is solid.
2. Shade each half-plane using the origin test $(0, 0)$ (legal here because the origin lies on none of the three boundary lines except $x = 0$; for $x \ge 0$ test a point such as $(1, 0)$ instead):
   • $2(0) + 0 = 0 \le 8$ is true, so shade the side of $y = -2x + 8$ containing the origin (below-left).
   • $0 - 0 = 0 < 2$ is true, so shade the side of $y = x - 2$ containing the origin (above-left).
   • $(1, 0)$ gives $1 \ge 0$, true, so shade to the right of the $y$-axis.
3. Identify the feasible region: the overlap of all three shaded half-planes — the wedge to the right of the $y$-axis, below $y = -2x + 8$, and above $y = x - 2$.
4. Solve boundary pairs to locate the corner vertices:
   • $x = 0$ with $y = -2x + 8$: vertex $(0, 8)$, included (both boundaries solid).
   • $x = 0$ with $y = x - 2$: vertex $(0, -2)$, not included, because $y = x - 2$ is dashed.
   • $y = -2x + 8$ with $y = x - 2$: set $-2x + 8 = x - 2 \implies 10 = 3x \implies x = \frac{10}{3}$, so $y = \frac{10}{3} - 2 = \frac{4}{3}$. Vertex $\left(\frac{10}{3}, \frac{4}{3}\right)$, not included.
5. Verify with an interior test point such as $(1, 1)$: $2(1) + 1 = 3 \le 8$ ✓, $1 - 1 = 0 < 2$ ✓, $1 \ge 0$ ✓. The point lies inside, confirming the region was shaded correctly.

Exam Trap: A strict inequality ($<$ or $>$) makes its boundary dashed and excludes every point on that line — including any corner vertex that sits on it. CLEP answer choices routinely differ only in whether an endpoint or vertex is included.