Graphing Linear Relationships — Solutions

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1. Gradient and y-intercept

   1. y = 3x + 5: m = 3, b = 5
   2. y = −2x + 7: m = −2, b = 7
   3. y = 4x − 1: m = 4, b = −1
   4. y = −x + 6: m = −1, b = 6
   5. y = 5x: m = 5, b = 0
   6. y = −3x − 4: m = −3, b = −4
   7. y = ½x + 3: m = ½, b = 3
   8. y = −⅔x + 1: m = −⅔, b = 1

2. Tables of values (x = −2, −1, 0, 1, 2)

   1. y = x + 4: y = 2, 3, 4, 5, 6
   2. y = 2x − 1: y = −5, −3, −1, 1, 3
   3. y = −x + 3: y = 5, 4, 3, 2, 1
   4. y = 3x + 2: y = −4, −1, 2, 5, 8
   5. y = −2x − 1: y = 3, 1, −1, −3, −5
   6. y = ½x + 2: y = 1, 1.5, 2, 2.5, 3

3. Sketch lines using gradient-intercept method

   1. y = x + 2: y-int (0,2); m=1 → (1,3), (2,4). Line rises left-to-right at 45°.
   2. y = 2x − 4: y-int (0,−4); m=2 → (1,−2), (2,0). x-int (2,0).
   3. y = −x + 3: y-int (0,3); m=−1 → (1,2), (2,1). x-int (3,0).
   4. y = −3x + 6: y-int (0,6); m=−3 → (1,3), (2,0). x-int (2,0).
   5. y = ½x − 1: y-int (0,−1); m=½ → (2,0), (4,1). x-int (2,0).
   6. y = 4: Horizontal line at y=4; gradient = 0; no x-intercept.

4. x-intercept and y-intercept method

   1. y = 2x − 6: y-int: (0, −6); x-int: set y=0 → x=3, so (3,0). Line passes through (0,−6) and (3,0).
   2. y = −3x + 9: y-int: (0, 9); x-int: set y=0 → x=3, so (3,0).
   3. 3x + 2y = 12: y-int: set x=0 → 2y=12, y=6, so (0,6). x-int: set y=0 → 3x=12, x=4, so (4,0).
   4. 4x − y = 8: y-int: set x=0 → −y=8, y=−8, so (0,−8). x-int: set y=0 → 4x=8, x=2, so (2,0).

5. Parallel lines

   1. Lines parallel to y = 3x + 1: (A) y = 3x − 5 and (C) y = 3x + 7 are parallel (both have m = 3).
   2. Parallel to y = −2x + 4 through y-int 6: y = −2x + 6
   3. How to identify parallel lines: Two lines are parallel if they have the same gradient (m value) but different y-intercepts.
   4. y = 4x − 3 and 2y = 8x + 10: Rearrange: y = 4x + 5. Both have m = 4, so yes, they are parallel.

6. Match equations to descriptions

   1. Rises steeply, y-int = 1: (i) y = 2x + 1 — m=2 (steep rise), b=1
   2. y-int = 4, gradient −1: (ii) y = −x + 4 — m=−1, b=4
   3. y-int = −2, gentle rise: (iii) y = ½x − 2 — m=½, b=−2
   4. y-int = −1, steep fall: (iv) y = −3x − 1 — m=−3, b=−1

7. Real-world linear models

   1. Plumber:
      1. C = 80h + 60: C = 80h + 60
      2. Gradient: m = 80; the plumber charges $80 for each additional hour of work.
      3. C-intercept: b = 60; the $60 call-out fee charged before any work begins.
      4. Sketch: Plot points: (0,60), (1,140), (2,220), (3,300), (4,380), (5,460). Straight line, C-axis start at 60.
      5. Hours for $380: 380 = 80h + 60 → 320 = 80h → h = 4 hours
   2. Car hire:
      1. Equation: C = 45d + 30
      2. 4-day cost: C = 45(4) + 30 = 180 + 30 = $210
      3. Days for $255: 255 = 45d + 30 → 225 = 45d → d = 5 days
   3. Cyclist d = 18t:
      1. Gradient: m = 18; the cyclist travels at 18 km/h.
      2. Distance in 2.5 h: d = 18 × 2.5 = 45 km
      3. Time for 126 km: 126 = 18t → t = 7 hours

8. Equation from two points

   1. (0,3) and (2,7): m = (7−3)/(2−0) = 2; b = 3.   y = 2x + 3
   2. (1,5) and (3,11): m = (11−5)/(3−1) = 3; sub (1,5): 5 = 3(1)+b → b=2.   y = 3x + 2
   3. (−1,4) and (2,1): m = (1−4)/(2−(−1)) = −3/3 = −1; sub (2,1): 1=−1(2)+b → b=3.   y = −x + 3
   4. (0,−2) and (4,6): m = (6−(−2))/(4−0) = 8/4 = 2; b = −2.   y = 2x − 2
   5. (2,−3) and (6,5): m = (5−(−3))/(6−2) = 8/4 = 2; sub (2,−3): −3=2(2)+b → b=−7.   y = 2x − 7
   6. (−2,7) and (4,−5): m = (−5−7)/(4−(−2)) = −12/6 = −2; sub (4,−5): −5=−2(4)+b → b=3.   y = −2x + 3

9. Graphing linear relationships in context (Water Tank)

   1. Equation: V = 15t + 20
   2. Table of values: t=0: V=20; t=1: V=35; t=2: V=50; t=3: V=65; t=4: V=80
   3. V-intercept: V-intercept = 20. This represents the 20 litres already in the tank before filling begins (at t = 0).
   4. Sketch: Plot points (0,20), (1,35), (2,50), (3,65), (4,80). Draw straight line. Label V-axis (litres), t-axis (minutes). Mark (0,20) as V-intercept.
   5. Time for 110 litres: 110 = 15t + 20 → 90 = 15t → t = 6 minutes
   6. Why straight line: The rate of change is constant — water is added at a fixed 15 L/min. A constant rate of change always produces a straight line (linear relationship).

10. Comparing two linear models (Taxi Services)

    1. Equations: Speedy Cabs: C = 3d + 5.   Quick Ride: C = 2d + 8.
    2. Sketch: Speedy Cabs: (0,5), (5,20), (10,35). Quick Ride: (0,8), (5,18), (10,28). Both straight lines on same axes for d = 0 to 10.
    3. Equal cost distance: 3d + 5 = 2d + 8 → d = 3 km. Both cost $14 at 3 km.
    4. Which is cheaper for d < 3 km: Quick Ride is cheaper for distances less than 3 km (lower starting cost). On the graph, the Quick Ride line sits below the Speedy Cabs line for d < 3.
    5. Cost for 7 km: Speedy Cabs: C = 3(7)+5 = $26. Quick Ride: C = 2(7)+8 = $22. Quick Ride is cheaper by $4.