2.3 Inequalities, Absolute Value, and Systems of Equations

2.3 Inequalities, Absolute Value, and Systems of Equations

Inequalities and absolute value ask a different question from the equations handled in the previous section: not "which single number works?" but "which set of numbers works?" Solutions are intervals and regions, and entrance-test items in this family tend to hide one procedural rule — a reversed symbol, a disjunction, an excluded case — inside otherwise routine algebra. Solving plain linear and quadratic equations, including the quadratic formula and discriminant, is covered separately; this section assumes you can already do that and focuses on what changes when the equals sign becomes an inequality or a system.

1. Linear Inequalities and the Sign-Flip Rule

Every operation that is legal on an equation is legal on an inequality, with one exception: multiplying or dividing both sides by a negative number reverses the direction of the symbol. Adding or subtracting a negative number does not reverse anything.

Worked Example 1

Solve $-3(x - 4) \ge 2x + 9$.

1. Distribute: $-3x + 12 \ge 2x + 9$.
2. Collect the variables on the left and the constants on the right: subtract $2x$ and $12$ from both sides, giving $-5x \ge -3$. Here that deliberately leaves a negative coefficient, which sets up the sign flip in the next step.
3. Divide by $-5$ and flip: $x \le \frac{3}{5}$.
4. Check with $x = 0$: the left side is $-3(0-4) = 12$ and the right side is $9$; since $12 \ge 9$ is true and $0 \le \frac{3}{5}$, the direction is right.

In interval notation the answer is $\left(-\infty, \frac{3}{5}\right]$.

Infinity always takes a parenthesis — you can never include it.

2. Compound Inequalities

An "and" (conjunction) statement keeps only the overlap; an "or" (disjunction) statement keeps everything in either piece. A three-part chain such as $-5 < 2x - 3 \le 7$ is an "and" written compactly: add 3 across all three parts to get $-2 < 2x \le 10$, then halve all three to get $-1 < x \le 5$, or $(-1, 5]$.

3. Absolute Value Equations and Inequalities

Absolute value measures distance from zero, so it is never negative. That single fact generates the whole table.

Solve $\lvert 3x - 4 \rvert = 11$ by splitting: $3x - 4 = 11$ gives $x = 5$, and $3x - 4 = -11$ gives $x = -\frac{7}{3}$.

For $\lvert x - 3 \rvert < 5$, rewrite as $-5 < x - 3 < 5$, then add 3 throughout: $-2 < x < 8$.

  |x - 3| < 5   ->   -2 < x < 8      (one connected interval)

  <---------o==================o--------->
           -2                  8

For $\lvert 2x + 1 \rvert \ge 9$, split into $2x + 1 \ge 9$, giving $x \ge 4$, and $2x + 1 \le -9$, giving $x \le -5$. The answer is $(-\infty, -5] \cup [4, \infty)$ — two rays, never a single chain.

  |2x + 1| >= 9  ->  x <= -5  or  x >= 4     (two separated rays)

  ==========*----------------------*==========>
           -5                      4

The mnemonic used across Philippine review classes is "less than = and, greater than = or".

4. Quadratic Inequalities by Sign Chart

Never divide an inequality by a variable expression — its sign is unknown. Move everything to one side, factor, and test signs.

Solve $x^2 \le 5x - 6$. Rearranged, $x^2 - 5x + 6 \le 0$, which factors as $(x-2)(x-3) \le 0$. The critical values 2 and 3 cut the number line into three regions:

                     2              3
   ------------------|--------------|------------------>
   (x - 2)      neg  |     pos      |      pos
   (x - 3)      neg  |     neg      |      pos
   product      POS  |     NEG      |      POS
                          ^ the only region where the product is <= 0

Because the inequality allows equality, the endpoints are included: $2 \le x \le 3$, or $[2, 3]$.

5. Systems of Linear Equations

Two lines in a plane either cross once, never cross, or coincide.

1. Substitution — isolate one variable and replace it in the other equation. Best when a coefficient is already 1.
2. Elimination — scale the equations so one variable cancels when you add or subtract.

Worked Example 2 — elimination

Solve $3x + 4y = 10$ and $5x - 2y = 8$. Multiply the second equation by 2 to get $10x - 4y = 16$, then add it to the first: $13x = 26$, so $x = 2$. Substituting back, $3(2) + 4y = 10$ gives $4y = 4$ and $y = 1$. Check the second original equation: $5(2) - 2(1) = 8$. The solution is $(2, 1)$.

Two special outcomes carry marks:

• Inconsistent (no solution). For $2x + 6y = 5$ and $x + 3y = 4$, doubling the second gives $2x + 6y = 8$; subtracting produces $0 = -3$, a false statement. The lines are parallel with different intercepts.
• Dependent (infinitely many solutions). For $y = 2x - 3$ and $4x - 2y = 6$, substitution gives $4x - 2(2x - 3) = 6$, which collapses to $6 = 6$, a statement true for every $x$. Both equations describe the same line.

6. Systems of Linear Inequalities

A system of inequalities defines a feasible region — the overlap of several half-planes. Graph each boundary line (solid for $\le$ or $\ge$, dashed for $<$ or $>$), then shade using a test point; $(0,0)$ is easiest whenever it is not on a boundary.

For $x \ge 0$, $y \ge 0$, $y \le -x + 6$ and $y \ge 2x - 3$, the point $(0,0)$ satisfies all four, so the region contains the origin. The corners are $(0,0)$, $(0,6)$, the intersection of the two slanted boundaries at $(3,3)$ — found by solving $-x + 6 = 2x - 3$ — and $(1.5, 0)$.

7. Translating Words into Inequalities

Two standard AdUCET-style setups:

• Minimum score. Ana scored 88, 79 and 92 on three 100-point quizzes and needs an average of at least 87 over four quizzes. Then $\frac{88+79+92+x}{4} \ge 87$, so $259 + x \ge 348$ and $x \ge 89$.
• Budget. A student has ₱600 for a ₱120 bag and notebooks at ₱45 each: $120 + 45n \le 600$ gives $45n \le 480$ and $n \le 10.67$. Because notebooks are whole objects, the answer is 10 — always round a counting answer down, whatever the decimal says.

Common Traps

• Flipping the symbol when you merely added or subtracted a negative number.
• Writing $\lvert x \rvert > 5$ as the impossible chain $-5 > x > 5$ instead of two separate rays.
• Forgetting that $\lvert u \rvert = c$ has no solution when $c$ is negative.
• Ignoring a check when the right side of an absolute-value equation contains a variable: $\lvert x - 2 \rvert = 3x$ yields $x = -1$ and $x = \frac{1}{2}$, but $x = -1$ makes the right side negative and must be rejected.
• Reading a collapse to $0 = 0$ as "$x = 0$" rather than as infinitely many solutions.
• Using only the critical values of a quadratic inequality without testing the sign of each region.