Practice Maths

Solutions — Solving Linear Equations and Inequalities

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  1. Q1 — Basic linear equations

    (a) 3x + 7 = 22

    Subtract 7 from both sides: 3x = 15

    Divide both sides by 3: x = 5

    Check: 3(5) + 7 = 15 + 7 = 22 ✓

    (b) 5x − 3 = 2x + 9

    Subtract 2x from both sides: 3x − 3 = 9

    Add 3 to both sides: 3x = 12

    Divide by 3: x = 4

    Check: LHS = 5(4)−3 = 17; RHS = 2(4)+9 = 17 ✓

    (c) 4(2x − 1) = 3(x + 5)

    Expand: 8x − 4 = 3x + 15

    Subtract 3x: 5x − 4 = 15

    Add 4: 5x = 19

    Divide by 5: x = 19/5 = 3.8

    Check: LHS = 4(2×3.8−1) = 4(6.6) = 26.4; RHS = 3(3.8+5) = 3(8.8) = 26.4 ✓

  2. Q2 — Equations with fractions

    (a) x/3 + 2 = 7

    Subtract 2: x/3 = 5

    Multiply both sides by 3: x = 15

    Check: 15/3 + 2 = 5 + 2 = 7 ✓

    (b) (2x + 1)/4 = 3

    Multiply both sides by 4: 2x + 1 = 12

    Subtract 1: 2x = 11

    Divide by 2: x = 5.5

    Check: (2×5.5+1)/4 = 12/4 = 3 ✓

    (c) x/2 − x/3 = 4

    LCM of 2 and 3 is 6. Multiply every term by 6:

    3x − 2x = 24

    x = 24

    Check: 24/2 − 24/3 = 12 − 8 = 4 ✓

  3. Q3 — Inequalities with number line

    (a) 2x − 3 > 7

    Add 3: 2x > 10

    Divide by 2: x > 5

    Number line: open circle at 5, arrow pointing right.

    (b) 5 − 3x ≤ 14

    Subtract 5: −3x ≤ 9

    Divide by −3 and flip the sign: x ≥ −3

    Number line: closed circle at −3, arrow pointing right.

    (c) 4x + 1 ≥ 2x − 5

    Subtract 2x: 2x + 1 ≥ −5

    Subtract 1: 2x ≥ −6

    Divide by 2: x ≥ −3

    Number line: closed circle at −3, arrow pointing right.

  4. Q4 — Equations with algebraic fractions

    (a) (x + 3)/2 = (2x − 1)/5

    Cross-multiply (multiply both sides by 10, the LCM of 2 and 5):

    5(x + 3) = 2(2x − 1)

    5x + 15 = 4x − 2

    Subtract 4x: x + 15 = −2

    Subtract 15: x = −17

    Check: LHS = (−17+3)/2 = −14/2 = −7; RHS = (2(−17)−1)/5 = −35/5 = −7 ✓

    (b) 3(x − 4)/2 + x = 8

    Multiply every term by 2 to clear the fraction:

    3(x − 4) + 2x = 16

    3x − 12 + 2x = 16

    5x − 12 = 16

    5x = 28

    x = 28/5 = 5.6

    Check: 3(5.6−4)/2 + 5.6 = 3(1.6)/2 + 5.6 = 2.4 + 5.6 = 8 ✓

  5. Q5 — Writing and solving: number problem

    Let the number be n.

    “A number is doubled and 7 is added, giving 29” → 2n + 7 = 29

    Subtract 7: 2n = 22

    Divide by 2: n = 11

    Check: 2(11) + 7 = 22 + 7 = 29 ✓

    The number is 11.

  6. Q6 — Writing and solving: inequality (books)

    Let n = number of books Alex buys.

    Cost of n books ≤ $85, so: 12n ≤ 85

    Divide both sides by 12: n ≤ 85/12 ≤ 7.083...

    Since n must be a whole number: n ≤ 7

    Alex can buy at most 7 books.

    Check: 7 × $12 = $84 ≤ $85 ✓   8 × $12 = $96 > $85 ✓

  7. Q7 — Multi-step equations

    (a) 2(3x − 4) = 5(x + 1) − 3

    Expand both sides: 6x − 8 = 5x + 5 − 3

    Simplify RHS: 6x − 8 = 5x + 2

    Subtract 5x: x − 8 = 2

    Add 8: x = 10

    Check: LHS = 2(30−4) = 52; RHS = 5(11)−3 = 55−3 = 52 ✓

    (b) (x − 2)/3 = (x + 4)/5

    Multiply both sides by 15 (LCM of 3 and 5):

    5(x − 2) = 3(x + 4)

    5x − 10 = 3x + 12

    Subtract 3x: 2x − 10 = 12

    Add 10: 2x = 22

    Divide by 2: x = 11

    Check: LHS = (11−2)/3 = 9/3 = 3; RHS = (11+4)/5 = 15/5 = 3 ✓

  8. Q8 — Real-world equation: plumber

    Let h = number of hours worked.

    Total cost = call-out fee + hourly rate × hours

    Equation: 65 + 85h = 322.50

    Subtract 65: 85h = 257.50

    Divide by 85: h = 257.50 ÷ 85 = 3.029... ≈ 3.03 hours

    Hmm — let’s verify: 65 + 85 × 3.03 ≈ 65 + 257.55 = 322.55. This is close but not exact due to rounding. Let’s compute exactly: 257.50 / 85 = 515/170 = 103/34 ≈ 3.029 hr.

    More naturally: 257.50 ÷ 85 = 3 hours and (257.50 − 255)/85 = 2.50/85 ≈ 0.029 hr (about 1.76 minutes). This suggests the problem intends 3 hours with a slight rounding in the bill, or the job took exactly 3 hours if the call-out fee was $67.50. Re-reading: 65 + 85×3 = 65 + 255 = $320 ≠ $322.50. Solving precisely: h = 257.50/85 = 53/17 ≈ 3.03 hours.

    The job took approximately 3 hours and 2 minutes (or the answer may be expressed as 53/17 hours ≈ 3.03 h).

  9. Q9 — Perimeter problem

    Let the width = w cm.

    Then the length = 2w + 8 cm (8 cm more than twice the width).

    Perimeter of a rectangle = 2(length + width) = 68

    2(2w + 8 + w) = 68

    2(3w + 8) = 68

    6w + 16 = 68

    6w = 52

    w = 52/6 = 26/3 ≈ 8.67 cm

    Length = 2(26/3) + 8 = 52/3 + 24/3 = 76/3 ≈ 25.33 cm

    Check: 2(76/3 + 26/3) = 2(102/3) = 2(34) = 68 ✓

    Width = 26/3 ≈ 8.67 cm; Length = 76/3 ≈ 25.33 cm

  10. Q10 — Age problem

    Let Sam’s current age = S, son’s current age = T.

    Equation 1 (Sam is 3 times as old as his son): S = 3T

    Equation 2 (in 10 years, Sam is twice as old as his son): S + 10 = 2(T + 10)

    Expand Equation 2: S + 10 = 2T + 20 → S = 2T + 10

    Substitute Equation 1 into this: 3T = 2T + 10

    Subtract 2T: T = 10

    So S = 3(10) = 30

    Sam is currently 30 years old; his son is 10 years old.

    Check: Sam is 3× as old ✓. In 10 years: Sam = 40, son = 20. 40 = 2×20 ✓