Inequalities belong to the exam's 20% algebra category; systems and their graphical interpretation cross into the 30% representations category — "how many solutions?" asked about a picture is a staple item type.
| Set | Interval | Bracket logic |
|---|---|---|
| x ≥ −5 | [−5, ∞) | square = included |
| x > −5 | (−5, ∞) | round = excluded |
| −3 < x < 4 | (−3, 4) | both endpoints out |
| x ≤ −7 or x ≥ 3 | (−∞, −7] ∪ [3, ∞) | union joins pieces |
Infinity always takes a round bracket — it is not a number to be included.
Solve like equations, with one exception: multiplying or dividing by a negative number reverses the inequality.
−3x + 5 ≤ 20 → −3x ≤ 15 → x ≥ −5 (divided by −3: flip)
Solution: [−5, ∞). The ≤ stays "or equal" through the flip — only the direction reverses, not the strictness.
To solve p(x) < 0 (or >, ≤, ≥):
Worked: x² − x − 6 < 0 → (x − 3)(x + 2) < 0. Zeros at −2 and 3.
| Interval | test x | (x − 3) | (x + 2) | product |
|---|---|---|---|---|
| x < −2 | −3 | − | − | + |
| −2 < x < 3 | 0 | − | + | − |
| x > 3 | 4 | + | + | + |
Negative between the roots: solution (−2, 3) — open, because the inequality is strict.
Repeated factors change the game. An even-power factor like (x − 2)² is ≥ 0 everywhere, so the polynomial does not change sign at x = 2 — the graph touches the axis and bounces. For (x + 1)(x − 2)² > 0: the sign is controlled by (x + 1), positive for x > −1, but x = 2 itself gives 0, which is not > 0. Solution: (−1, 2) ∪ (2, ∞) — the bounce point is punched out.
[GRAPH: y = (x + 1)(x − 2)² in the xy-plane. Curve crosses the x-axis at x = −1 rising left-to-right, then touches the axis at x = 2 (local minimum on the axis) and rises again. Region above the x-axis shaded for x > −1 with an open hole at x = 2. Caption: "even multiplicity: touch, no sign change — the zero itself fails a strict inequality."]
For (x + 1)/(x − 4) ≥ 0, the sign can change at zeros of the numerator (x = −1) and zeros of the denominator (x = 4). Build the chart with both:
| Interval | test x | sign |
|---|---|---|
| x < −1 | −2 | (−)/(−) = + |
| −1 < x < 4 | 0 | (+)/(−) = − |
| x > 4 | 5 | (+)/(+) = + |
Endpoint rules differ: x = −1 makes the expression 0, allowed by ≥ → included. x = 4 makes it undefined → excluded, always, no matter the inequality symbol. Solution: (−∞, −1] ∪ (4, ∞).
Never multiply both sides by an expression containing x. For 3/(x − 2) > 1, multiplying by (x − 2) silently assumes it's positive; when x < 2 the direction should flip. Correct route — move everything to one side:
3/(x − 2) − 1 > 0 → (3 − (x − 2))/(x − 2) > 0 → (5 − x)/(x − 2) > 0
Boundary points 2 and 5; sign chart gives positive only between them: (2, 5). The shortcut "3 > x − 2, so x < 5" wrongly includes everything below 2.
| Form | Rewrite | Shape |
|---|---|---|
| |X| < c | −c < X < c | one bounded interval ("AND") |
| |X| > c | X > c or X < −c | two rays ("OR") |
Less-than traps you inside a band; greater-than throws you outside it. Worked:
Writing the second as −5 ≥ x + 2 ≥ 5 (a "sandwich" for a greater-than) is self-contradictory — no number is both. Match the shape to the symbol before writing anything.
Substitution (one equation easily solved for a variable) and elimination (align coefficients, add) both work; pick per problem. Worked by substitution:
2x + 3y = 12, x − y = 1 → x = 1 + y
2(1 + y) + 3y = 12 → 2 + 5y = 12 → y = 2, x = 3
Check in both originals: 2(3) + 3(2) = 12 ✓ and 3 − 2 = 1 ✓.
Three geometric outcomes for two lines:
| Algebra says | Geometry | Solutions |
|---|---|---|
| one (x, y) pair | lines cross | exactly one |
| false statement (0 = 2) | parallel, distinct | none |
| true statement (0 = 0) | same line | infinitely many |
Spot parallels fast: 3x − 6y = 9 is x − 2y = 3 after dividing by 3 — same left side as x − 2y = 5, different right side. Same slope, different intercept: no solution.
Line-parabola: substitute the line into the quadratic and solve.
y = x² − 2x and y = x + 4:
x² − 2x = x + 4 → x² − 3x − 4 = 0 → (x − 4)(x + 1) = 0 → x = 4, −1
Back-substitute into the line for y: (4, 8) and (−1, 3). A solution of a system is a point, not just an x-value — finish the job.
Line-circle: same substitution idea. x² + y² = 25 with y = x + 1 gives x² + (x + 1)² = 25 → 2x² + 2x − 24 = 0 → x² + x − 12 = 0 → x = 3 or x = −4 → two intersection points, (3, 4) and (−4, −3).
Graphical solution counts — the geometry caps the algebra:
[GRAPH: three panels, each a parabola opening upward with a line. Panel 1: line misses the parabola entirely — 0 solutions. Panel 2: line tangent, touching at exactly one point — 1 solution. Panel 3: line cutting through both branches — 2 solutions. Caption: "solution count = intersection count."]
Q1. C. −3x ≤ 15; dividing by −3 flips the inequality: x ≥ −5, i.e., [−5, ∞). A) solves correctly but forgets to flip, keeping x ≤ −5. B) flips correctly but uses a round bracket — ≤ keeps the endpoint. D) drops the sign of −5 during the division. E) combines the dropped sign with the unflipped direction. Fix: flip the inequality exactly when you multiply or divide by a negative; the "or equal" part rides along unchanged.
Q2. A. (x − 3)(x + 2) < 0 between the roots: test x = 0 gives (−3)(2) = −6 < 0 ✓, so (−2, 3). B) selects the outside intervals, where the product is positive — that solves > 0. C) includes the endpoints, but a strict < excludes the zeros. D) uses the factor constants with flipped signs as interval endpoints — the roots are −2 and 3, not −3 and 2. E) makes both errors of B and C. Fix: for a positive-leading-coefficient quadratic, "< 0" lives between the roots, with open endpoints when strict.
Q3. A. Sign chart boundaries at −1 (numerator zero) and 4 (denominator zero): positive on (−∞, −1) and (4, ∞); x = −1 gives 0, allowed by ≥, so it's included; x = 4 is undefined, so excluded. B) includes x = 4 — but a denominator zero can never be in the solution set, whatever the symbol. C) picks the interval where the expression is negative. D) same wrong interval, opened. E) drops x = −1, which the ≥ legitimately includes. Fix: numerator zeros follow the inequality symbol; denominator zeros are excluded unconditionally.
Q4. C. |2x − 1| < 7 → −7 < 2x − 1 < 7 → −6 < 2x < 8 → −3 < x < 4. A) uses the "OR" (greater-than) shape for a less-than — inside-the-band and outside-the-band swapped. B) mishandles the +1 step, adding 1 with the wrong sign to each side. D) closes the endpoints, but the strict < keeps them out. E) solves only the right half of the sandwich. Fix: |X| < c is a single bounded interval — write the three-part inequality first, then operate on all three parts at once.
Q5. B. |x + 2| ≥ 5 → x + 2 ≥ 5 or x + 2 ≤ −5 → x ≥ 3 or x ≤ −7. A) uses the "AND" band shape, which belongs to |X| ≤ c. C) subtracts the 2 with the wrong sign in both cases, getting −3 and 7. D) combines the shape error with the sign error. E) uses round brackets, but ≥ includes both endpoints. Fix: greater-than means OR — two rays pointing away from each other, endpoints included exactly when the symbol has the "or equal."
Q6. B. From x − y = 1, x = 1 + y; substituting: 2(1 + y) + 3y = 12 → 5y = 10 → y = 2, so x = 3. Both checks pass: 6 + 6 = 12, 3 − 2 = 1. A) reports y instead of x — read what the question asks. C) adds x + y = 5 and stops, reporting the sum. D) reports the right side of the second equation. E) solves 2x = 12 as if y vanished, halving 12. Fix: after solving a system, label which value is x and which is y, and answer the variable actually requested.
Q7. E. Set x² − 2x = x + 4: x² − 3x − 4 = 0 → (x − 4)(x + 1) = 0 → x = 4, −1; then y from the line: y = 8 and y = 3, giving (4, 8) and (−1, 3). A) and B) each keep only one root of the quadratic. C) pairs each x with the other solution's y — coordinates must come from substituting that x. D) drops the sign on the second root, using x = 1 (which gives y = 5 on the line but does not lie on the parabola: 1 − 2 = −1 ≠ 5). Fix: solve the combined quadratic for all x-values, then compute each y by substituting its own x into one original equation.
Q8. D. Substituting a line into a quadratic yields a quadratic (or linear, for vertical lines) equation with at most 2 real roots: 0 (miss), 1 (tangent), or 2 (secant) intersection points — all three occur. A) forgets that lines can miss or cut twice. B) omits the tangent case, where the discriminant is zero. C) allows 3, but a quadratic cannot have three roots. E) unbounded counts require the "curves" to coincide, impossible for a line and a parabola. Fix: the number of intersections equals the number of real roots of the substituted equation — for a quadratic, the discriminant's sign gives 0, 1, or 2.
Q9. D. Dividing the first equation by 3 gives x − 2y = 3, which contradicts x − 2y = 5; subtracting yields 0 = 2, a false statement — parallel distinct lines, no solution. A) is the generic expectation, but these slopes match. B) two distinct lines can never cross twice. C) would need the equations to be the same line (0 = 0 after elimination), but the right sides differ. E) the equations fully determine the answer — elimination settles it. Fix: when elimination produces a false constant statement, the lines are parallel and the system has no solution; a true statement like 0 = 0 means infinitely many.
Q10. D. Substitute y = x + 1: x² + (x + 1)² = 25 → 2x² + 2x − 24 = 0 → x² + x − 12 = 0 → (x + 4)(x − 3) = 0 — two real x-values, hence two intersection points, (3, 4) and (−4, −3). A) assumes the line misses, but the discriminant 1 + 48 = 49 > 0. B) is the tangent case, needing discriminant zero. C) and E) exceed what a line and circle allow — at most 2. Fix: substitute the line into the circle and count real roots of the resulting quadratic; a line meets a circle in at most two points.
Q11. C. (x − 2)² ≥ 0 always, so the product's sign follows (x + 1): positive exactly when x > −1 — except x = 2, where the product is 0, and 0 > 0 fails. Solution (−1, 2) ∪ (2, ∞). A) forgets to remove the bounce point x = 2 from a strict inequality. B) picks the side where (x + 1) is negative. D) treats x = 2 as a sign-change crossing and keeps only the far-right interval. E) stops at the touch point, as if the product turned negative beyond it. Fix: even-multiplicity zeros don't change sign — for strict inequalities, delete those isolated zeros from the solution with a union.
Q12. E. Multiplying by (x − 2) is only valid if its sign is known; for x < 2 the direction should flip, so the student's set wrongly includes everything below 2. Correct method: 3/(x − 2) − 1 > 0 → (5 − x)/(x − 2) > 0, positive exactly on (2, 5). Spot check: x = 0 gives 3/(−2) = −1.5, not > 1, so x < 5 alone is wrong. A) accepts the flawed step and its inflated answer. B) endpoint tinkering doesn't repair the missing case split. C) reverses the interval — values above 5 make the fraction less than 1 (x = 8: 3/6 = 0.5). D) unions the two regions where the expression is negative or small, the opposite of what's wanted. Fix: never multiply an inequality by a variable expression; move everything to one side, combine into a single fraction, and run a sign chart on numerator and denominator zeros.
Q1. C. −3x ≤ 15; dividing by −3 flips the inequality: x ≥ −5, i.e., [−5, ∞). A) solves correctly but forgets to flip, keeping x ≤ −5. B) flips correctly but uses a round bracket — ≤ keeps the endpoint. D) drops the sign of −5 during the division. E) combines the dropped sign with the unflipped direction. Fix: flip the inequality exactly when you multiply or divide by a negative; the "or equal" part rides along unchanged.
Q2. A. (x − 3)(x + 2) < 0 between the roots: test x = 0 gives (−3)(2) = −6 < 0 ✓, so (−2, 3). B) selects the outside intervals, where the product is positive — that solves > 0. C) includes the endpoints, but a strict < excludes the zeros. D) uses the factor constants with flipped signs as interval endpoints — the roots are −2 and 3, not −3 and 2. E) makes both errors of B and C. Fix: for a positive-leading-coefficient quadratic, "< 0" lives between the roots, with open endpoints when strict.
Q3. A. Sign chart boundaries at −1 (numerator zero) and 4 (denominator zero): positive on (−∞, −1) and (4, ∞); x = −1 gives 0, allowed by ≥, so it's included; x = 4 is undefined, so excluded. B) includes x = 4 — but a denominator zero can never be in the solution set, whatever the symbol. C) picks the interval where the expression is negative. D) same wrong interval, opened. E) drops x = −1, which the ≥ legitimately includes. Fix: numerator zeros follow the inequality symbol; denominator zeros are excluded unconditionally.
Q4. C. |2x − 1| < 7 → −7 < 2x − 1 < 7 → −6 < 2x < 8 → −3 < x < 4. A) uses the "OR" (greater-than) shape for a less-than — inside-the-band and outside-the-band swapped. B) mishandles the +1 step, adding 1 with the wrong sign to each side. D) closes the endpoints, but the strict < keeps them out. E) solves only the right half of the sandwich. Fix: |X| < c is a single bounded interval — write the three-part inequality first, then operate on all three parts at once.
Q5. B. |x + 2| ≥ 5 → x + 2 ≥ 5 or x + 2 ≤ −5 → x ≥ 3 or x ≤ −7. A) uses the "AND" band shape, which belongs to |X| ≤ c. C) subtracts the 2 with the wrong sign in both cases, getting −3 and 7. D) combines the shape error with the sign error. E) uses round brackets, but ≥ includes both endpoints. Fix: greater-than means OR — two rays pointing away from each other, endpoints included exactly when the symbol has the "or equal."
Q6. B. From x − y = 1, x = 1 + y; substituting: 2(1 + y) + 3y = 12 → 5y = 10 → y = 2, so x = 3. Both checks pass: 6 + 6 = 12, 3 − 2 = 1. A) reports y instead of x — read what the question asks. C) adds x + y = 5 and stops, reporting the sum. D) reports the right side of the second equation. E) solves 2x = 12 as if y vanished, halving 12. Fix: after solving a system, label which value is x and which is y, and answer the variable actually requested.
Q7. E. Set x² − 2x = x + 4: x² − 3x − 4 = 0 → (x − 4)(x + 1) = 0 → x = 4, −1; then y from the line: y = 8 and y = 3, giving (4, 8) and (−1, 3). A) and B) each keep only one root of the quadratic. C) pairs each x with the other solution's y — coordinates must come from substituting that x. D) drops the sign on the second root, using x = 1 (which gives y = 5 on the line but does not lie on the parabola: 1 − 2 = −1 ≠ 5). Fix: solve the combined quadratic for all x-values, then compute each y by substituting its own x into one original equation.
Q8. D. Substituting a line into a quadratic yields a quadratic (or linear, for vertical lines) equation with at most 2 real roots: 0 (miss), 1 (tangent), or 2 (secant) intersection points — all three occur. A) forgets that lines can miss or cut twice. B) omits the tangent case, where the discriminant is zero. C) allows 3, but a quadratic cannot have three roots. E) unbounded counts require the "curves" to coincide, impossible for a line and a parabola. Fix: the number of intersections equals the number of real roots of the substituted equation — for a quadratic, the discriminant's sign gives 0, 1, or 2.
Q9. D. Dividing the first equation by 3 gives x − 2y = 3, which contradicts x − 2y = 5; subtracting yields 0 = 2, a false statement — parallel distinct lines, no solution. A) is the generic expectation, but these slopes match. B) two distinct lines can never cross twice. C) would need the equations to be the same line (0 = 0 after elimination), but the right sides differ. E) the equations fully determine the answer — elimination settles it. Fix: when elimination produces a false constant statement, the lines are parallel and the system has no solution; a true statement like 0 = 0 means infinitely many.
Q10. D. Substitute y = x + 1: x² + (x + 1)² = 25 → 2x² + 2x − 24 = 0 → x² + x − 12 = 0 → (x + 4)(x − 3) = 0 — two real x-values, hence two intersection points, (3, 4) and (−4, −3). A) assumes the line misses, but the discriminant 1 + 48 = 49 > 0. B) is the tangent case, needing discriminant zero. C) and E) exceed what a line and circle allow — at most 2. Fix: substitute the line into the circle and count real roots of the resulting quadratic; a line meets a circle in at most two points.
Q11. C. (x − 2)² ≥ 0 always, so the product's sign follows (x + 1): positive exactly when x > −1 — except x = 2, where the product is 0, and 0 > 0 fails. Solution (−1, 2) ∪ (2, ∞). A) forgets to remove the bounce point x = 2 from a strict inequality. B) picks the side where (x + 1) is negative. D) treats x = 2 as a sign-change crossing and keeps only the far-right interval. E) stops at the touch point, as if the product turned negative beyond it. Fix: even-multiplicity zeros don't change sign — for strict inequalities, delete those isolated zeros from the solution with a union.
Q12. E. Multiplying by (x − 2) is only valid if its sign is known; for x < 2 the direction should flip, so the student's set wrongly includes everything below 2. Correct method: 3/(x − 2) − 1 > 0 → (5 − x)/(x − 2) > 0, positive exactly on (2, 5). Spot check: x = 0 gives 3/(−2) = −1.5, not > 1, so x < 5 alone is wrong. A) accepts the flawed step and its inflated answer. B) endpoint tinkering doesn't repair the missing case split. C) reverses the interval — values above 5 make the fraction less than 1 (x = 8: 3/6 = 0.5). D) unions the two regions where the expression is negative or small, the opposite of what's wanted. Fix: never multiply an inequality by a variable expression; move everything to one side, combine into a single fraction, and run a sign chart on numerator and denominator zeros.