2.4 Non-Linear Systems and Absolute Value Equations

2.4 Non-Linear Systems and Absolute Value Equations

Absolute value functions and non-linear systems introduce non-linear geometry and multi-case branching logic. This section establishes the algebraic properties required to solve absolute value equations, compound inequalities, and non-linear system intersections on the CLEP Precalculus exam.

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1. Absolute Value Equations

The absolute value of a real number $x$, denoted $|x|$, represents the non-negative distance between $x$ and $0$ on a real number line:

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

Solving Basic Absolute Value Equations

For any algebraic expression $u$ and constant $C$:

1. If $C > 0$, then $|u| = C \iff u = C \quad \text{or} \quad u = -C$.
2. If $C = 0$, then $|u| = 0 \iff u = 0$.
3. If $C < 0$, then $|u| = C$ has no real solution ($\emptyset$).

Variable Right-Hand Side Equations ($|u| = g(x)$)

When an absolute value equation equals a variable expression $|u(x)| = g(x)$, split the equation into two cases ($u = g(x)$ or $u = -g(x)$) and verify candidate solutions in the original equation. Any solution resulting in $g(x) < 0$ is extraneous because an absolute value cannot equal a negative number.

Worked Example 1: Absolute Value Equation with Extraneous Root

Solve the equation for $x$:

$|2x - 5| = 3x + 1$

Solution:

1. Set up the two algebraic cases:
   • Case 1: $2x - 5 = 3x + 1$ $2x - 3x = 1 + 5 \implies -x = 6 \implies x = -6$
   • Case 2: $2x - 5 = -(3x + 1)$ $2x - 5 = -3x - 1 \implies 5x = 4 \implies x = \frac{4}{5}$
2. Verify candidate solutions in the original equation $|2x - 5| = 3x + 1$:
   • Check $x = -6$: $|2(-6) - 5| = |-12 - 5| = |-17| = 17$ Right-hand side: $3(-6) + 1 = -18 + 1 = -17$. Since $17 \neq -17$, the value $x = -6$ is extraneous.
   • Check $x = \frac{4}{5}$: $\left|2\left(\frac{4}{5}\right) - 5\right| = \left|\frac{8}{5} - \frac{25}{5}\right| = \left|-\frac{17}{5}\right| = \frac{17}{5}$ Right-hand side: $3\left(\frac{4}{5}\right) + 1 = \frac{12}{5} + \frac{5}{5} = \frac{17}{5}$.
     Since $\frac{17}{5} = \frac{17}{5}$, $x = \frac{4}{5}$ is valid.
3. The unique solution is $x = \frac{4}{5}$.

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2. Absolute Value Inequalities

Absolute value inequalities translate into compound linear inequalities based on distance concepts.

Fundamental Absolute Value Inequality Rules ($c > 0$)

1. Less Than / Less Than or Equal To (AND Condition - Bounded Interval): $|u| < c \iff -c < u < c$ $|u| \le c \iff -c \le u \le c$
2. Greater Than / Greater Than or Equal To (OR Condition - Unbounded Union): $|u| > c \iff u > c \quad \text{or} \quad u < -c$ $|u| \ge c \iff u \ge c \quad \text{or} \quad u \le -c$

Negative Constant Cases ($c < 0$)

• $|u| < -5 \implies \text{No Solution } (\emptyset)$ (distance cannot be negative).
• $|u| > -5 \implies \text{All Real Numbers } (-\infty, \infty)$ (absolute value is always $\ge 0 > -5$).

Worked Example 2: Absolute Value Inequality with Union

Solve the inequality and express the solution set in interval notation:

$|4 - 3x| \ge 7$

Solution:

1. Apply the OR rule for $\ge$: $4 - 3x \ge 7 \quad \text{or} \quad 4 - 3x \le -7$
2. Solve the first inequality: $4 - 3x \ge 7 \implies -3x \ge 3 \implies x \le -1$
3. Solve the second inequality: $4 - 3x \le -7 \implies -3x \le -11 \implies x \ge \frac{11}{3}$
4. Combine into interval notation: $(-\infty, -1] \cup \left[\frac{11}{3}, \infty\right)$

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3. Systems of Non-Linear Equations

A non-linear system of equations contains at least one equation that is not linear (e.g., quadratics, circles, parabolas). The solutions to the system represent the physical $(x, y)$ coordinate points where the graphs intersect.

Number of Possible Intersections

• Line and Circle/Parabola: $0$, $1$ (tangent), or $2$ intersection points.
• Two Conic Sections (e.g., Circle and Ellipse): Up to $4$ distinct intersection points.

Algebraic Solution Strategies

• Substitution Method: Best when one equation can be solved easily for $x$ or $y$ (e.g., a linear equation in a line-conic system).
• Elimination Method: Best when both equations contain matching squared terms (e.g., $x^2$ and $y^2$).

Worked Example 3: System of a Line and a Circle

Find all points of intersection for the system:

$\begin{cases} x^2 + y^2 = 25 \quad \text{--- (Eq. 1: Circle)} \\ y - x = 1 \qquad \text{--- (Eq. 2: Line)} \end{cases}$

Solution:

1. Solve (Eq. 2) for $y$: $y = x + 1$
2. Substitute $y = x + 1$ into (Eq. 1): $x^2 + (x + 1)^2 = 25$
3. Expand and simplify: $x^2 + (x^2 + 2x + 1) = 25 \implies 2x^2 + 2x + 1 = 25$ $2x^2 + 2x - 24 = 0$
4. Divide by $2$ and factor: $x^2 + x - 12 = 0 \implies (x + 4)(x - 3) = 0$ $x = -4 \quad \text{or} \quad x = 3$
5. Find corresponding $y$-values using $y = x + 1$:
   • If $x = -4 \implies y = -4 + 1 = -3$. Intersection point: $(-4, -3)$.
   • If $x = 3 \implies y = 3 + 1 = 4$. Intersection point: $(3, 4)$.
6. The system has two intersection points: $(-4, -3)$ and $(3, 4)$.

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4. Absolute Value & Non-Linear Reference Table