A22: Inequalities

Solve: 3x + 2 > 11

Solution:

Subtract 2: 3x > 9

Divide by 3: x > 3

Solve: 5x - 3 ≤ 2x + 9

Solution:

Subtract 2x: 3x - 3 ≤ 9

Add 3: 3x ≤ 12

Divide by 3: x ≤ 4

Solve: -2x < 8

Solution:

Divide by -2 and reverse sign: x > -4

Remember: when dividing by negative, flip the sign!

Represent x > 2 on a number line.

Solution: Open circle at 2, arrow pointing right (towards greater values).

Represent -3 ≤ x < 5 on a number line.

Solution: Filled circle at -3, open circle at 5, line between them.

List the integer solutions of -2 ≤ x < 4

Solution:

x can be -2, -1, 0, 1, 2, 3

(Note: x cannot be 4 because of the strict inequality)

Write x ≥ 1 in set notation.

Solution: {x : x ≥ 1}

Solve: x² - 3x - 4 > 0

Solution:

Step 1: Find roots: x² - 3x - 4 = 0

(x - 4)(x + 1) = 0

x = 4 or x = -1

Step 2: Sketch parabola (opens upward, a > 0)

Step 3: We want where y > 0 (above x-axis)

Answer: x < -1 or x > 4

Solve: x² + 2x - 15 ≤ 0

Solution:

Roots: (x + 5)(x - 3) = 0

x = -5 or x = 3

Parabola opens upward. We want y ≤ 0 (below or on x-axis)

Answer: -5 ≤ x ≤ 3

Solve 5x + 3 ≥ 2x − 9

Collect x terms on one side, numbers on the other:

5x − 2x ≥ −9 − 3  →  3x ≥ −12

Divide by 3 (positive — no flip): x ≥ −4

M1 for valid rearrangement, A1 for x ≥ −4. Writing the answer as an equation (x = −4) loses the final mark — keep the symbol.

List the integer values that satisfy −2 ≤ x < 3

x can equal −2 (filled circle — ≤) but not 3 (open circle — <).

Answer: −2, −1, 0, 1, 2

B1 all five values. Sketching the number line first, with the circles, makes the boundary values obvious.

The perimeter of a rectangle is less than 30 cm. Its length is x cm and its width is 4 cm. Write an inequality for x and solve it.

Perimeter = 2(x + 4)  →  2(x + 4) < 30

Divide both sides by 2: x + 4 < 15

Subtract 4: x < 11

M1 for forming 2(x + 4) < 30, A1 for x < 11. 'Form and solve' inequality questions also award a mark for correct context — lengths must also satisfy x > 0.

Q1. 4x − 7 > 13 → 4x > 20 (add 7) → x > 5 (divide by 4)

Q2. 2x + 5 ≤ x + 8 → x ≤ 3 (subtract x, subtract 5)

Q3. −3x > 12 → divide by −3 and FLIP: x < −4. Forgetting the flip is the single most common error on this topic.

Q4. −3 < x ≤ 2 → open at −3, filled at 2 → −2, −1, 0, 1, 2

Q5. x² − 9 > 0 → (x + 3)(x − 3) > 0. Sketch y = x²: above the x-axis means x < −3 or x > 3 → x < −3 or x > 3

Q6. x² + 5x + 6 < 0 → (x + 2)(x + 3) < 0 → between the roots → −3 < x < −2

Q7. 7 − 2x ≥ 3 → −2x ≥ −4 → divide by −2, FLIP → x ≤ 2

Q8. 4x − 4 < 2x + 6 → 2x < 10 → x < 5