Inequality Fundamentals
Solving Inequalities Algebraically
Compound & Quadratic Inequalities
Both conditions must hold. The solution is the overlap.
Either condition can hold. The solution is both regions.
Number Line & Visual Range Logic
10 Inequality Traps
1. FLIP the sign when multiplying/dividing by negative
$-2x > 4$ → $x < -2$. The most common algebra error on inequalities.
2. Don't multiply by an unknown sign
If you don't know if $x$ is positive or negative, you can't multiply both sides by $x$ and flip based on assumption.
3. Squaring both sides: only safe if both sides are non-negative
$x > y$ does NOT imply $x^2 > y^2$ if either side could be negative.
4. Testing DS inequalities: try multiple values
In DS problems with inequalities, test positive, negative, and zero values.
5. "Greater than zero" test for DS
$\frac{x}{y} > 0$ is true when x and y have the SAME sign. Not just when x > 0.
6. Compound inequality — must maintain all parts
$-3 < 2x+1 < 7$ means solve as a block: subtract 1 from all three parts, divide by 2.
7. $x^2 < 9$ means $-3 < x < 3$
NOT $x < 3$. The solution includes negative values.
8. Consecutive integers under inequality
$n < x < n+2$ where $x$ is integer means $x = n+1$. Count carefully.
9. $x > 0$ does NOT mean $x$ is an integer
Fractional values are also valid unless stated otherwise.
10. Absolute value inequalities: two cases
$|x| < 3$ → $-3 < x < 3$. $|x| > 3$ → $x < -3$ or $x > 3$.
10 GMAT Practice Questions
Solve for $x$: $3x - 7 > 8$.
(C) $x > 5$. $3x > 15$ → $x > 5$.
Solve for $x$: $-4x \geq 20$.
(D) $x \leq -5$. Divide by $-4$ (negative) → flip the sign. $x \leq -5$.
If $x^2 < 16$, which of the following must be true?
(C) $-4 < x < 4$. $x^2 < 16$ means $|x| < 4$, which means $-4 < x < 4$.
Solve: $-3 \leq 2x + 1 \leq 9$.
(A) $-2 \leq x \leq 4$. Subtract 1 from all: $-4 \leq 2x \leq 8$. Divide by 2: $-2 \leq x \leq 4$.
For what values of $x$ is $(x-1)(x+3) > 0$?
(B) $x < -3$ or $x > 1$. Critical points: $x = 1$ and $x = -3$. The product is positive when both factors have the same sign: both negative ($x<-3$) or both positive ($x>1$).
Is $x > 0$?
(1) $x^2 > 0$
(2) $x^3 > 0$
(B) Statement (2) alone sufficient. (1): $x^2 > 0$ means $x \neq 0$, but $x$ could be negative. Not sufficient. (2): $x^3 > 0$ only when $x > 0$ (odd power preserves sign). Sufficient. Answer: (B).
If $x$ and $y$ are both positive and $x > y$, which of the following must be true?
(C) $x^2 > y^2$. Both positive and $x > y$ → $x^2 > y^2$ (squaring is monotone for positive values). Other options: (A) false, $x/y > 1$. (E) false: larger denominator means smaller fraction.
What is the integer solution set for $|2x - 3| < 5$?
(B) {−1, 0, 1, 2, 3}. $|2x-3|<5$ → $-5 < 2x-3 < 5$ → $-2 < 2x < 8$ → $-1 < x < 4$. Integer solutions: $\{0, 1, 2, 3\}$. Wait: $-1 < x < 4$ → integers: 0,1,2,3. But $-1$ is excluded. So (A) $\{0,1,2,3\}$ is correct. (A) $\{0,1,2,3\}$.
If $a < b < 0$, which of the following is greatest?
(D) $b - a$. Both $a$ and $b$ are negative with $a < b$. Example: $a=-5, b=-2$. $a-b = -3$. $b-a = 3$ (positive!). $a+b=-7$. $b-a$ is the greatest since it's the only positive value.
Is $\dfrac{a}{b} > 0$?
(1) $a \cdot b < 0$
(2) $a > b$
(A) Statement (1) alone sufficient. (1): $ab < 0$ means $a$ and $b$ have OPPOSITE signs. Thus $\frac{a}{b} < 0$. Definitively NO. Sufficient. (2): $a > b$ tells us ordering but not signs. Both could be positive or one negative. NOT sufficient. Answer: (A).
Lesson Summary — Key Takeaways
Flip sign for negative division
Dividing by −4: $-4x ≥ 20$ → $x ≤ -5$. The inequality reverses.
$x^2 < a^2$ means $-a < x < a$
The quadratic inequality gives a closed interval between $-a$ and $a$.
Quadratics: factor and test regions
Find critical points, test one value per region, mark which regions satisfy the inequality.
DS inequality: test positive, negative, zero
Always verify by plugging in different value types — never assume the variable's sign.