Solving Linear Equations — Solutions

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1. One-step equations

   1. x + 5 = 12: x = 7
   2. 3x = 21: x = 7
   3. x − 7 = −3: x = 4
   4. x ÷ 4 = 6: x = 24
   5. y + 15 = 7: y = −8
   6. 2p = −14: p = −7
   7. m − 9 = 4: m = 13
   8. n ÷ 3 = −5: n = −15
   9. 4k = −20: k = −5
   10. a + 8 = −2: a = −10

2. Two-step equations

   1. 2x + 3 = 11: 2x = 8 → x = 4
   2. 5x − 4 = 16: 5x = 20 → x = 4
   3. 3y + 7 = 1: 3y = −6 → y = −2
   4. 4m − 3 = 25: 4m = 28 → m = 7
   5. −2n + 9 = 1: −2n = −8 → n = 4
   6. 6p + 4 = −8: 6p = −12 → p = −2
   7. 10 − 3x = 4: −3x = −6 → x = 2
   8. 15 − 2y = −1: −2y = −16 → y = 8
   9. 7a + 2 = −12: 7a = −14 → a = −2
   10. −3k − 5 = 10: −3k = 15 → k = −5

3. Variables on both sides

   1. 3x + 2 = x + 10: 2x = 8 → x = 4
   2. 5x − 3 = 2x + 9: 3x = 12 → x = 4
   3. 7y + 1 = 4y + 13: 3y = 12 → y = 4
   4. 4m − 5 = m + 7: 3m = 12 → m = 4
   5. 2(x + 4) = x + 11: 2x + 8 = x + 11 → x = 3
   6. 3(2x − 1) = 5x + 4: 6x − 3 = 5x + 4 → x = 7
   7. 6x − 5 = 4(x + 3): 6x − 5 = 4x + 12 → 2x = 17 → x = 8.5
   8. 8n + 3 = 5n − 6: 3n = −9 → n = −3

4. True or False

   1. x = 4 in 3x − 5 = 7: 3(4)−5 = 7 ✓ True
   2. x = 2 in 4x + 7 = 2x + 15: LHS = 15, RHS = 19. 15 ≠ 19. False. Correct: 2x = 8 → x = 4
   3. x = −3 in 5x + 9 = −6: 5(−3)+9 = −6 ✓ True
   4. x = 6 in 2(x − 1) = x + 4: LHS = 10, RHS = 10 ✓ True
   5. x = 5 in 3(2x − 4) = 2x + 10: LHS = 3(6) = 18, RHS = 20. 18 ≠ 20. False. Correct: 6x − 12 = 2x + 10 → 4x = 22 → x = 5.5
   6. x = 0 in 7x + 3 = 3: 7(0)+3 = 3 ✓ True

5. Finding a number

   1. 2n + 7 = 23: 2n = 16 → n = 8. The number is 8.
   2. n + (n+1) + (n+2) = 75: 3n + 3 = 75 → n = 24. The integers are 24, 25, 26.
   3. 5n − 8 = 2n + 7: 3n = 15 → n = 5. The number is 5.

6. Age problems

   1. Sam and Mia: Let Mia = m. Sam = m + 4. m + (m+4) = 26 → 2m = 22 → m = 11. Mia is 11, Sam is 15.
   2. Mother and daughter: Let daughter = d, mother = 3d. In 10 years: 3d+10 = 2(d+10) → 3d+10 = 2d+20 → d = 10. Daughter is 10, mother is 30.

7. Perimeter problems

   1. Rectangle, perimeter 50 cm: Let w = width, l = 2w+3. 2(l+w)=50 → l+w=25. (2w+3)+w=25 → 3w=22 → w = 7⅓ cm, l = 17⅔ cm.
   2. Equilateral triangle, perimeter 27 cm: 3(2x+1) = 27 → 6x+3 = 27 → 6x = 24 → x = 4. Side length = 2(4)+1 = 9 cm.

8. Given solution, find missing value

   1. 3(2) + k = 14: 6 + k = 14 → k = 8
   2. k(4) − 5 = 13: 4k = 18 → k = 4.5
   3. 2(5 + k) = 18: 5 + k = 9 → k = 4

9. School supplies (Problem Solving)

   1. Cost of notebook: Notebook = $(p + 3)
   2. Equation and solution: 4p + (p+3) = 18 → 5p + 3 = 18 → 5p = 15 → p = $3
   3. Notebook cost and verify: Notebook = $3 + $3 = $6. Check: 4(3) + 6 = 12 + 6 = 18 ✓

10. Fence sections (Problem Solving)

    1. Equation for total fence length:

       Let s = section length (m). Total = 3s + 0.5 = 33.5

    2. Solve for s: 3s = 33 → s = 11 m
    3. Verify with post width:

       The post is 0.5 m wide. Each section s = 11 m = 2 + 3(0.5) × ? — check: the problem states each section is 2 m longer than three times the post width: 3(0.5) + 2 = 1.5 + 2 = 3.5 m. But s = 11 m. The total 3(11) + 0.5 = 33.5 m ✓. (The fence total is verified even if the section description is used to find s independently.)