Substitution Method — Solutions

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1. Direct substitution

   1. y = x + 3, 2x + y = 12: Sub: 2x + (x+3) = 12 → 3x = 9 → x = 3, y = 6
   2. y = 3x, x + y = 8: x + 3x = 8 → 4x = 8 → x = 2, y = 6
   3. y = 4x − 1, x + y = 9: x + (4x−1) = 9 → 5x = 10 → x = 2, y = 7
   4. y = 2x + 5, 3x + y = 20: 3x + (2x+5) = 20 → 5x = 15 → x = 3, y = 11
   5. y = 5 − x, 2x + y = 7: 2x + (5−x) = 7 → x = 2, y = 3
   6. y = x − 4, 3x − y = 16: 3x − (x−4) = 16 → 2x + 4 = 16 → x = 6, y = 2
   7. x = 2y, x + 3y = 10: 2y + 3y = 10 → 5y = 10 → y = 2, x = 4
   8. x = y + 1, 2x + y = 11: 2(y+1) + y = 11 → 3y = 9 → y = 3, x = 4
   9. y = 3x − 2, 2x + y = 13: 2x + (3x−2) = 13 → 5x = 15 → x = 3, y = 7
   10. y = 6 − 2x, x + 2y = 7: x + 2(6−2x) = 7 → x + 12 − 4x = 7 → −3x = −5 → x = 5/3, y = 8/3

2. Rearrange then substitute

   1. x + y = 9, 3x + 2y = 22: From (1): y = 9−x. Sub: 3x+2(9−x)=22 → x=4, y=5
   2. 2x + y = 10, 5x + 3y = 27: From (1): y=10−2x. Sub: 5x+3(10−2x)=27 → −x=−3 → x=3, y=4
   3. x − y = 2, 4x + y = 23: From (1): x=y+2. Sub: 4(y+2)+y=23 → 5y=15 → y=3, x=5
   4. 3x − y = 4, x + 2y = 12: From (1): y=3x−4. Sub: x+2(3x−4)=12 → 7x=20 → x=20/7 … Try from eq2: x=12−2y. Sub: 3(12−2y)−y=4 → 36−7y=4 → y=32/7, x=20/7
   5. x + 2y = 11, 2x + 3y = 18: From (1): x=11−2y. Sub: 2(11−2y)+3y=18 → 22−y=18 → y=4, x=3
   6. 4x + y = 17, 2x + 3y = 11: From (1): y=17−4x. Sub: 2x+3(17−4x)=11 → 2x+51−12x=11 → −10x=−40 → x=4, y=1

3. Check solutions

   1. y=2x−1, x+y=8; claimed x=3, y=5: Check eq1: 5 = 2(3)−1 = 5 ✓   eq2: 3+5=8 ✓. Correct solution.
   2. y=x+4, 2x+y=13; claimed x=3, y=7: Check eq1: 7=3+4=7 ✓   eq2: 2(3)+7=13 ✓. Correct solution.
   3. y=3x+1, x+y=9; claimed x=2, y=6: Check eq1: 6=3(2)+1=7 ✗ Incorrect. Correct: x+3x+1=9 → 4x=8 → x=2, y=7

4. Word problems

   1. Two numbers summing to 18, one 4 more: Let x+y=18, x=y+4. Sub: y+4+y=18 → 2y=14 → y=7, x=11
   2. Pen and pencil: Let p=pen, c=pencil. p=c+2, p+c=5.50. Sub: c+2+c=5.50 → 2c=3.50 → c=$1.75, p=$3.75
   3. 24 students, girls = 2 × boys: Let g+b=24, g=2b. Sub: 2b+b=24 → 3b=24 → b=8 boys, g=16 girls
   4. Amara and Josh, $30 difference, $110 total: Let A=J+30, A+J=110. Sub: J+30+J=110 → 2J=80 → J=$40, A=$70
   5. Rectangle perimeter 36 cm, l=2w+3: 2l+2w=36 → l+w=18. Sub l=2w+3: 2w+3+w=18 → 3w=15 → w=5 cm, l=13 cm
   6. School fair tickets, 7 tickets $44: Let a+c=7 and 8a+5c=44. From (1): c=7−a. Sub: 8a+5(7−a)=44 → 3a=9 → a=3 adults, c=4 children

5. Word problems with equation setup

   1. Cinema tickets ($12 adult, $8 student, 30 tickets, $296): Let a+s=30, 12a+8s=296. From (1): s=30−a. Sub: 12a+8(30−a)=296 → 4a=56 → a=14, s=16. 14 adult, 16 student.
   2. Chickens and cows (20 animals, 56 legs): Let c+k=20, 4c+2k=56 (cows have 4 legs, chickens 2). From (1): k=20−c. Sub: 4c+2(20−c)=56 → 2c=16 → c=8 cows, k=12 chickens. 8 cows, 12 chickens.
   3. Two taps, A is 3 L/min faster, together 21 L/min: Let A=B+3, A+B=21. Sub: B+3+B=21 → 2B=18 → B=9 L/min, A=12 L/min.
   4. Two numbers, difference 7, 2×larger + smaller = 40: Let x−y=7 (x larger), 2x+y=40. From (1): x=y+7. Sub: 2(y+7)+y=40 → 3y=26 → y=26/3 → not integer — recheck: 2x+y=40 with x=y+7: 2y+14+y=40 → 3y=26 → y=8⅔, x=15⅔. x ≈ 15.67, y ≈ 8.67 (or exact: x=47/3, y=26/3).

6. Geometry with simultaneous equations

   1. Supplementary angles, one = 5 × other: x+y=180, x=5y. Sub: 5y+y=180 → 6y=180 → y=30°, x=150°. 30° and 150°.
   2. Rectangle perimeter 50, length = 4+3×width: 2l+2w=50 → l+w=25, l=3w+4. Sub: 3w+4+w=25 → 4w=21 → w=5.25 cm, l=19.75 cm. Area=5.25×19.75=103.6875 cm².
   3. Isosceles triangle, perimeter 32, y=x−8: 2x+y=32, y=x−8. Sub: 2x+(x−8)=32 → 3x=40 → x=40/3 — recheck: 2x+y=32, y=x−8 → 2x+x−8=32 → 3x=40 → x=13⅓ cm, y=5⅓ cm. Equal sides = 40/3 cm, base = 16/3 cm.

7. Equations with fractions

   1. y=½x+3, x+y=9: Sub: x+(½x+3)=9 → &frac32;x=6 → x=4, y=5. x=4, y=5
   2. y=⅓x−2, 2x+y=11: Sub: 2x+(⅓x−2)=11 → &frac73;x=13 → x=39/7 … try integer check: 2x+⅓x=13 → &frac73;x=13 → x=13×3/7=39/7 ≈ 5.57. y=13/7−2=−1/7. x=39/7, y=−1/7
   3. x+y=5, 3x−2y=25: From (1): y=5−x. Sub: 3x−2(5−x)=25 → 5x=35 → x=7, y=−2. x=7, y=−2
   4. 2x−3y=1, x=2y−1: Sub: 2(2y−1)−3y=1 → 4y−2−3y=1 → y=3, x=5. x=5, y=3

8. Consistent, inconsistent and dependent systems

   1. y=2x+1, 4x−2y=−2: Sub: 4x−2(2x+1)=−2 → 4x−4x−2=−2 → −2=−2. Identity ⇒ infinitely many solutions (same line).
   2. y=3x+4, y=3x−1: Sub: 3x+4=3x−1 → 4=−1. Contradiction ⇒ no solution (parallel lines).
   3. y=−2x+5, 2x+y=5: Sub: 2x+(−2x+5)=5 → 5=5. Identity ⇒ infinitely many solutions (same line).

9. Rate and mixture problems

   1. Cyclist and runner, opposite directions, 48 km in 1 hr: Let c=cyclist speed, r=runner speed. c=3r, c+r=48. Sub: 3r+r=48 → r=12 km/h, c=36 km/h. Runner: 12 km/h; Cyclist: 36 km/h.
   2. Coffee blend, $18/kg and $12/kg, 5 kg for $78: Let a+b=5, 18a+12b=78. From (1): b=5−a. Sub: 18a+12(5−a)=78 → 6a=18 → a=3 kg Type A, b=2 kg Type B. 3 kg Type A, 2 kg Type B.
   3. 3x+2y=29, x=y+1: Sub: 3(y+1)+2y=29 → 5y=26 → y=26/5 — recheck: 3y+3+2y=29 → 5y=26 → y=5.2, x=6.2. x=6.2, y=5.2 (or exact: x=31/5, y=26/5).

10. Extended challenge

    1. Bank accounts X and Y:
       1. Account X: Pr=80. Account Y: (P+200)r=110.
       2. From (1): r=80/P. Sub into (2): (P+200)(80/P)=110 → 80+16000/P=110 → 16000/P=30 → P=16000/30=1600/3 ≈ $533.33. r=80÷(1600/3)=80×3/1600=0.15. P ≈ $533.33, r=0.15
       3. 15% per year.
    2. Line y=ax+b through (2,7) and (5,16):
       1. 2a+b=7 and 5a+b=16.
       2. From (1): b=7−2a. Sub: 5a+(7−2a)=16 → 3a=9 → a=3, b=1. y=3x+1
       3. y=3(10)+1=31