Practice Maths

★ Topic Review — Linear Equations and Their Graphs — Solutions

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This review covers all four lessons: Solving Linear Equations and Inequalities, Graphing Linear Functions, Linear Models and Applications, and Simultaneous Equations. Questions increase in difficulty.

  1. Fluency — Solving linear equations and inequalities

    1. (a) Solve: 5x − 8 = 2x + 7
    2. (b) Solve: (3x + 1)/4 = (x − 2)/2
    3. (c) Solve the inequality and represent on a number line: 3 − 2x > 11

    (a) 5x − 8 = 2x + 7 → 3x = 15 → x = 5

    (b) Multiply both sides by 4: 3x + 1 = 2(x − 2) = 2x − 4 → x = −5. x = −5

    Check: LHS = (3(−5)+1)/4 = −14/4 = −3.5; RHS = (−5−2)/2 = −7/2 = −3.5 ✓

    (c) 3 − 2x > 11 → −2x > 8 → divide by −2 and flip: x < −4

    Number line: open circle at −4, arrow pointing left.

  2. Fluency — Features of a linear function

    For the line y = −4x + 8:

    1. (a) State the gradient m and y-intercept c.
    2. (b) Find the x-intercept.
    3. (c) Find the y-value when x = −3.
    4. (d) Find the x-value when y = −4.

    (a) m = −4, c = 8.

    (b) Set y = 0: 0 = −4x + 8 → 4x = 8 → x = 2. x-intercept: (2, 0).

    (c) y = −4(−3) + 8 = 12 + 8 = 20

    (d) −4 = −4x + 8 → −4x = −12 → x = 3

  3. Fluency — Simultaneous equations

    1. (a) Solve by substitution: 2x + 3y = 15   and   x − y = 0
    2. (b) Solve by elimination: 3x + 2y = 16   and   3x − y = 7

    (a) From Eq (2): x = y. Substitute into Eq (1): 2y + 3y = 15 → 5y = 15 → y = 3. x = 3. Solution: x = 3, y = 3.

    (b) Subtract Eq (2) from Eq (1): (3x + 2y) − (3x − y) = 16 − 7 → 3y = 9 → y = 3. Substitute: 3x + 6 = 16 → 3x = 10 → x = 10/3 ≈ 3.33. Solution: x = 10/3, y = 3.

  4. Understanding — Equation of a line

    Find the equation of the line through (−2, 5) with gradient −3.

    1. (a) Write in gradient–intercept form y = mx + c.
    2. (b) Write in general form ax + by + c = 0 (with integer coefficients).

    (a) y − 5 = −3(x − (−2)) → y − 5 = −3x − 6 → y = −3x − 1

    (b) Rearrange: 3x + y + 1 = 0. General form: 3x + y + 1 = 0

  5. Understanding — Parallel, perpendicular, or same line?

    Are the lines y = 3x − 1 and 2y = 6x + 4 parallel, perpendicular, or the same line? Explain fully.

    Rearrange the second equation: y = 3x + 2.

    Line 1: m = 3, c = −1.   Line 2: m = 3, c = 2.

    Same gradient (m = 3) but different y-intercepts (−1 ≠ 2), so the lines never intersect.

    The lines are parallel.

  6. Understanding — Perpendicular line

    Find the equation of the line perpendicular to y = 4x − 3 that passes through (8, 1).

    Original gradient m = 4. Perpendicular gradient m⊥ = −1/4.

    Using point (8, 1): y − 1 = −¼(x − 8) → y − 1 = −¼x + 2 → y = −¼x + 3

    Check: at (8,1): y = −2 + 3 = 1 ✓

  7. Understanding — Linear model: mobile data

    A mobile plan includes 10 GB of data. Each additional GB costs $5.

    1. (a) Write a cost model C for extra data, in terms of e (GB used over the 10 GB limit).
    2. (b) Find the extra cost if the total data used in a month is 13.6 GB.

    (a) Extra data e = total usage − 10 GB (only charged when e > 0).

    Rate = $5/GB, no extra fixed charge for overuse: C = 5e   (for e ≥ 0)

    (b) Extra data = 13.6 − 10 = 3.6 GB. C = 5(3.6) = $18

  8. Understanding — Multi-step equation

    Solve: 4(x − 2) − 3(2x + 1) = −5

    Expand: 4x − 8 − 6x − 3 = −5

    Collect: −2x − 11 = −5

    −2x = 6 → x = −3

    Check: 4(−3−2) − 3(2(−3)+1) = 4(−5) − 3(−5) = −20 + 15 = −5 ✓

  9. Understanding — Simultaneous equations: ticket sales

    Adult tickets cost $22 each and child tickets cost $14 each. A group of 12 people paid a total of $204. How many adults and how many children were in the group?

    Let a = number of adults, c = number of children.

    Eq (1): a + c = 12

    Eq (2): 22a + 14c = 204

    From Eq (1): a = 12 − c. Substitute into Eq (2):

    22(12 − c) + 14c = 204 → 264 − 22c + 14c = 204 → −8c = −60 → c = 7.5

    This is not a whole number — let’s double-check: 22a + 14c = 204 with a + c = 12. If c = 6: a = 6, total = 22(6) + 14(6) = 132 + 84 = 216 ≠ 204. If c = 9: a = 3, total = 66 + 126 = 192 ≠ 204. If c = 7: a = 5, total = 110 + 98 = 208 ≠ 204. If c = 8: a = 4, total = 88 + 112 = 200 ≠ 204.

    Solving exactly: −8c = 204 − 264 = −60 → c = 7.5. Since this is not an integer, the problem likely intends a slightly different total. Using the algebra as written: c = 7.5 adults and a = 4.5 children is not realistic. The answer the problem intends using standard rounding: the simultaneous solution is a = 4.5, c = 7.5. The nearest integer solution: 5 adults and 7 children (gives $110 + $98 = $208) or 4 adults and 8 children (gives $88 + $112 = $200).

    Note to student: check your arithmetic on this type of problem carefully, as a decimal answer usually signals an error in reading the problem or an error in the calculation.

    Using the exact algebra: a = 12 − 7.5 = 4.5, c = 7.5. Since this problem is designed to have whole number answers, the most likely intended values are: change the total to $208 for 5 adults and 7 children, or $200 for 4 adults and 8 children. As written with $204: the algebraic solution is a = 4.5, c = 7.5. This is a well-posed algebra exercise even if the numbers are not whole.

  10. Understanding — Spring model

    A spring is 10 cm at rest and stretches 2 cm for every 1 N of force applied.

    1. (a) Write the equation for length L (cm) in terms of force F (N).
    2. (b) Find the length at F = 6 N.
    3. (c) What force would cause a length of 22 cm?

    (a) c = 10 (natural length), m = 2 cm/N. L = 2F + 10

    (b) L = 2(6) + 10 = 12 + 10 = 22 cm

    (c) 22 = 2F + 10 → 2F = 12 → F = 6 N

  11. Understanding — Parallel line from general form

    Find the equation of the line through (2, −3) that is parallel to 4x − 2y = 10.

    Rearrange 4x − 2y = 10: y = 2x − 5. Gradient m = 2.

    Parallel line also has m = 2. Using point (2, −3):

    y − (−3) = 2(x − 2) → y + 3 = 2x − 4 → y = 2x − 7

    Check: (2, −3): y = 4 − 7 = −3 ✓

  12. Problem Solving — Inequality with algebraic fractions

    Solve the inequality (x + 3)/2 ≥ (2x − 1)/3 and represent the solution on a number line.

    Multiply both sides by 6 (LCM of 2 and 3):

    3(x + 3) ≥ 2(2x − 1)

    3x + 9 ≥ 4x − 2

    Subtract 4x: −x + 9 ≥ −2

    Subtract 9: −x ≥ −11

    Multiply by −1 and flip: x ≤ 11

    Number line: closed circle at 11, arrow pointing left.

    Check (x = 11): LHS = 14/2 = 7; RHS = 21/3 = 7. Equal ✓ (boundary point).

    Check (x = 9): LHS = 12/2 = 6; RHS = 17/3 ≈ 5.67. 6 ≥ 5.67 ✓

  13. Problem Solving — Comparing car hire companies

    Luxury Cars: $120/day + $0.35/km.   Budget Wheels: $75/day + $0.52/km.

    1. (a) Write cost equations for each company for one day in terms of km travelled (k).
    2. (b) Find the km per day that makes both companies equal cost.
    3. (c) Which company is cheaper for a 3-day trip with 150 km/day? How much is saved?

    (a) Luxury: C_L = 0.35k + 120   Budget: C_B = 0.52k + 75

    (b) Set equal: 0.35k + 120 = 0.52k + 75

    45 = 0.17k → k = 45/0.17 ≈ 264.7 km per day

    Both cost the same at approximately 265 km/day.

    (c) At 150 km/day (which is less than 265 km break-even → Budget should be cheaper):

    Luxury daily: 0.35(150) + 120 = 52.50 + 120 = $172.50

    Budget daily: 0.52(150) + 75 = 78 + 75 = $153.00

    3-day Luxury total: 3 × $172.50 = $517.50

    3-day Budget total: 3 × $153.00 = $459.00

    Budget Wheels is cheaper by $517.50 − $459.00 = $58.50 over the 3-day trip.

  14. Problem Solving — Finding k

    A straight line passes through (1, k) and (k, 9) and has gradient 2.

    1. (a) Use the gradient formula to write an equation in k and solve for k.
    2. (b) Write the coordinates of both points and find the equation of the line.

    (a) Gradient formula: m = (9 − k)/(k − 1) = 2

    9 − k = 2(k − 1)

    9 − k = 2k − 2

    11 = 3k → k = 11/3

    Hmm, that gives a non-integer result. Let’s check: could k be an integer? Let’s try k = 3: gradient = (9−3)/(3−1) = 6/2 = 3 ≠ 2. k = 4: gradient = (9−4)/(4−1) = 5/3 ≠ 2. k = 11/3 is exact.

    Verify: gradient = (9 − 11/3)/(11/3 − 1) = (27/3 − 11/3)/(11/3 − 3/3) = (16/3)/(8/3) = 2 ✓

    k = 11/3

    (b) Points: (1, 11/3) and (11/3, 9).

    Using m = 2 and point (1, 11/3):

    y − 11/3 = 2(x − 1) → y = 2x − 2 + 11/3 = 2x + 5/3

    y = 2x + 5/3

    Check with (11/3, 9): y = 2(11/3) + 5/3 = 22/3 + 5/3 = 27/3 = 9 ✓

  15. Problem Solving — Resource allocation

    A factory produces widgets (x) and gadgets (y).

    Each widget uses 4 kg metal + 2 hr labour.
    Each gadget uses 3 kg metal + 5 hr labour.
    Available per day: 48 kg metal, 45 hr labour.

    Solve the simultaneous equations to find how many widgets and gadgets should be produced per day to use all available metal and labour.

    Set up the system:

    Metal: 4x + 3y = 48   … (1)

    Labour: 2x + 5y = 45   … (2)

    Elimination — eliminate x: Multiply Eq (2) by 2: 4x + 10y = 90

    Subtract Eq (1): (4x + 10y) − (4x + 3y) = 90 − 48 → 7y = 42 → y = 6

    Substitute y = 6 into Eq (1): 4x + 18 = 48 → 4x = 30 → x = 7.5

    The factory should produce 7.5 widgets and 6 gadgets per day to use all resources.

    Check metal: 4(7.5) + 3(6) = 30 + 18 = 48 ✓

    Check labour: 2(7.5) + 5(6) = 15 + 30 = 45 ✓

    Note: A non-integer answer (7.5 widgets) is mathematically correct for this system, though in practice the factory might need to produce either 7 or 8 widgets and adjust slightly.