9.3 First-Degree Linear Systems, Simple Inequalities (Basit Eşitsizlikler) & Absolute Value (Mutlak Değer)

Linear Systems, Simple Inequalities (Basit Eşitsizlikler) & Absolute Value (Mutlak Değer)

---

First-Degree Linear Equations & $2 \times 2$ Systems (Birinci Dereceden Denklemler)

1. Parametric Single-Variable Equations ($ax + b = 0$)

The solution set (çözüm kümesi, $ÇK$) of $ax + b = 0$ over $\mathbb{R}$ depends entirely on whether the coefficient of $x$ vanishes:

• Unique Solution (Tek Elemanlı Çözüm Kümesi): $a \ne 0 \implies x = -\frac{b}{a}$.
• Empty Solution Set ($ÇK = \varnothing$): $a = 0$ and $b \ne 0$ (reduces to the contradiction $0 \cdot x + b = 0$).
• Infinite Solutions ($ÇK = \mathbb{R}$): $a = 0$ and $b = 0$ (reduces to the identity $0 = 0$, satisfied by every $x \in \mathbb{R}$).

2. Two-Variable Linear Systems (İki Bilinmeyenli Denklem Sistemleri)

For the system of two lines $a_1 x + b_1 y = c_1$ and $a_2 x + b_2 y = c_2$:

$$\begin{cases} \text{1. Unique Intersection Point:} & \frac{a_1}{a_2} \ne \frac{b_1}{b_2} \\ \text{2. Infinite Solutions (Coincident Lines):} & \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \\ \text{3. Empty Solution Set } \varnothing \text{ (Parallel Lines):} & \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \end{cases}$$

---

Simple Inequalities (Basit Eşitsizlikler) & Sign Traps

1. Core Inequality Operations & Reciprocals

• Multiplying or dividing both sides of an inequality by a negative number reverses the inequality sign: if $a < b$ and $c < 0$, then $a c > b c$.
• Reciprocal Rule: If $a$ and $b$ have the same sign ($0 < a < b$ or $a < b < 0$), taking reciprocals reverses the inequality sign: $\frac{1}{a} > \frac{1}{b}$. If $a$ and $b$ have opposite signs ($a < 0 < b$), taking reciprocals preserves the sign: $\frac{1}{a} < 0 < \frac{1}{b}$.

3. Squaring an Interval That Crosses Zero

• Same-sign bounds: If $2 < x < 5$, then $4 < x^2 < 25$. If $-5 < x \le -2$, squaring reverses the order: $4 \le x^2 < 25$.
• Opposite-sign bounds (crossing $0$): If $-a < x < b$ with $a, b > 0$, then $x = 0$ lies inside the interval, so the minimum possible value of $x^2$ is $0^2 = 0$! The upper bound is $\max(a^2, b^2)$:

$$-4 < x < 3 \implies 0 \le x^2 < 16 \qquad (\text{NEVER } 9 < x^2 < 16!)$$

4. The Domain Trap: Integer Selection ($\mathbb{Z}$) vs. Real Interval Expansion ($\mathbb{R}$)

Before touching an inequality range question, check the opening declaration of the variables:

• If $x, y \in \mathbb{Z}$ (Integers): Never multiply and add the inequalities! Pick discrete extreme integers directly from each interval. For example, if $x, y \in \mathbb{Z}$ with $-2 < x < 4$ ($x \in \{-1, 0, 1, 2, 3\}$) and $1 < y < 6$ ($y \in \{2, 3, 4, 5\}$), the maximum value of $2x - 3y$ uses the largest integer $x = 3$ and smallest integer $y = 2$: $2(3) - 3(2) = 0$.
• If $x, y \in \mathbb{R}$ (Real Numbers, or no integer condition is stated): You must expand the continuous intervals algebraically: $-4 < 2x < 8$ and $-18 < -3y < -3$, which sum to $-22 < 2x - 3y < 5$, so the maximum integer value of $2x - 3y$ is $4$!
• Quadratic Range Trap on $\mathbb{R}$: To bound $x^2 + 2bx + c$ on an interval of $x \in \mathbb{R}$, never bound $x^2$ and $2bx$ separately and add them (because $x$ cannot attain two different values simultaneously). Always complete the square as $(x + b)^2 + (c - b^2)$ first, then shift and square the interval for $x$.

---

Absolute Value (Mutlak Değer): Equations, Inequalities & Extrema

1. Piecewise Sign-Peeling (Mutlak Değerden Çıkarma)

To remove absolute value bars from $\lvert E \rvert$, determine the sign of the interior expression $E$:

$$\lvert E \rvert = \begin{cases} E, & E \ge 0 \\ -E, & E < 0 \end{cases}$$

For example, if $a < 0 < b$, then $a - b < 0 \implies \lvert a - b \rvert = -(a - b) = b - a$, while $b - a > 0 \implies \lvert b - a \rvert = b - a$. Also use scaling and symmetry: $\lvert -x \rvert = \lvert x \rvert$, $\lvert a - b \rvert = \lvert b - a \rvert$, and $\lvert k x - k a \rvert = \lvert k \rvert \cdot \lvert x - a \rvert$.

2. Absolute Value Equations & The Extraneous Root Trap

• Constant Right Side ($\lvert f(x) \rvert = c$ with $c > 0$): Splits into $f(x) = c$ or $f(x) = -c$.
• Two Absolute Values ($\lvert f(x) \rvert = \lvert g(x) \rvert$): Splits into $f(x) = g(x)$ or $f(x) = -g(x)$.
• Variable Right Side ($\lvert f(x) \rvert = g(x)$) — Extraneous Root Trap: Because $\lvert f(x) \rvert \ge 0$, any valid root must satisfy $g(x) \ge 0$. Solve both $f(x) = g(x)$ and $f(x) = -g(x)$, then substitute each root back into $g(x) \ge 0$ and discard any root where $g(x) < 0$!

3. Absolute Value Inequalities & Triangle Inequality Extrema

• Upper Bound ($\lvert f(x) \rvert \le k$ for $k \ge 0$): Equivalent to the single sandwich interval $-k \le f(x) \le k$.
• Lower Bound ($\lvert f(x) \rvert \ge k$): Equivalent to the two-ray union $f(x) \ge k$ or $f(x) \le -k$.
• Sum & Difference Extrema of Two Absolute Values ($a \ne b$):
  • Sum $S(x) = \lvert x - a \rvert + \lvert x - b \rvert$: Measures the sum of distances from $x$ to $a$ and $b$. Its minimum value is $\lvert a - b \rvert$, attained for every $x$ between $a$ and $b$. (It has no maximum on $\mathbb{R}$.)
  • Difference $D(x) = \lvert x - a \rvert - \lvert x - b \rvert$: Satisfies $-\lvert a - b \rvert \le \lvert x - a \rvert - \lvert x - b \rvert \le \lvert a - b \rvert$. Thus its maximum value is $+\lvert a - b \rvert$ and its minimum value is $-\lvert a - b \rvert$.

---

Summary Table of Systems, Inequalities & Absolute Value Rules

---

Worked Examples

Worked Example 1: Quadratic Range Over a Zero-Crossing Real Interval

Problem: Given that $x \in \mathbb{R}$ and $-2 < x \le 4$, find the interval of all possible values of $A = x^2 - 2x + 3$.

Solution:

1. Because $x \in \mathbb{R}$ appears in both $x^2$ and $-2x$, complete the square first:

$$A = (x^2 - 2x + 1) + 2 = (x - 1)^2 + 2$$

1. Shift the given interval $-2 < x \le 4$ by subtracting $1$:

$$-3 < x - 1 \le 3$$

1. Since $(-3, 3]$ crosses $0$, squaring sets the minimum to $0^2 = 0$ and the maximum to $3^2 = 9$ (included because $x - 1 = 3$ is included):

$$0 \le (x - 1)^2 \le 9 \xrightarrow{+2} 2 \le (x - 1)^2 + 2 \le 11$$

Thus $A$ lies in the closed interval $[2, 11]$.

Worked Example 2: Filtering Extraneous Roots in $\lvert f(x) \rvert = g(x)$

Problem: Find the real solution set of $\lvert 2x - 3 \rvert = x + 3$.

Solution:

1. Impose the non-negativity condition on the right side: $x + 3 \ge 0 \implies x \ge -3$.
2. Solve both linear branches:
   • Branch 1: $2x - 3 = x + 3 \implies x = 6$ (satisfies $6 \ge -3$, valid).
   • Branch 2: $2x - 3 = -(x + 3) = -x - 3 \implies 3x = 0 \implies x = 0$ (satisfies $0 \ge -3$, valid). Both roots pass the right-hand sign check, so the solution set is $ÇK = \{0, 6\}$. (By contrast, if the right side had been $x - 6$, requiring $x \ge 6$, the candidate roots $x = -3$ and $x = 3$ would both fail $x \ge 6$, leaving $ÇK = \varnothing$!)

Linear Systems Without Division by Zero

The displayed ratio tests assume their denominators are nonzero and each equation defines a line. A safer general method is elimination. Compute $D=a_1b_2-a_2b_1$. If $D\ne0$, there is one solution. If $D=0$, compare proportional coefficient rows and constants: a proportional full row gives the same equation, while mismatched constants give a contradiction. Treat rows such as $0x+0y=0$ as unrestricted and $0x+0y=c$ with $c\ne0$ as impossible. This also handles horizontal and vertical lines.

Piecewise Absolute-Value Equations Can Have Intervals of Answers

Consider $|x-1|+|x-5|=8$. The sign-change points are 1 and 5, so divide the number line into three regions before removing the absolute-value bars. For $x<1$, the left side is $(1-x)+(5-x)=6-2x$, which equals 8 at $x=-1$. This candidate belongs to the region and is valid. Between 1 and 5, the expression is $(x-1)+(5-x)=4$, so no point in that interval solves the equation with right side 8. For $x>5$, it is $(x-1)+(x-5)=2x-6$, giving the valid solution $x=7$. Substitution confirms both values.

Change the right side to 4 and the solution changes qualitatively: every real number in the closed interval $[1,5]$ works, including the endpoints. If the problem instead restricts $x$ to integers, there are five solutions: 1, 2, 3, 4 and 5. Thus an absolute-value equation need not have only one or two isolated roots. Always retain the domain and test the boundary points rather than relying on the apparent shape of an equation.

Review the Method

By the triangle inequality, the minimum of $\lvert x - a \rvert + \lvert x - b \rvert$ is $\lvert a - b \rvert$, and the extrema of $\lvert x - a \rvert - \lvert x - b \rvert$ are $\pm \lvert a - b \rvert$.