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2a Solving Equations Bar Method.pkdkdmptx

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Monday 3 March 2025 Solving Equations with the Bar Method Starte r C W
VALUE OF A FUNCTION Monday 3 March 202 5 Solving Equations with the Bar Method
One Step Equations
π‘₯+5=20 20 5 What is the value of ?
What is the value of ? π‘Ž βˆ’ 25=35 35 25
π‘₯+25=35 10=π‘₯+8 π‘₯ βˆ’6=18 10=π‘₯ βˆ’2 80 +π‘₯=60 10=20 βˆ’ π‘₯ Can you represent these using a bar method. What is the value of ? 35 25 10 8 60 80 18 6 10 2 10
2 𝑣=2 8 What is the value of ? 28
π‘š Γ· 4=6 What is the value of ? π’Ž 6 6 6 6
5 π‘₯=35 80=10 π‘₯ 18 Γ· π‘₯=3 π‘₯ Γ· 4=5 Can you represent these using a bar method. What is the value of ? 35 π‘₯ π‘₯ π‘₯ π‘₯ π‘₯ 80 π‘₯π‘₯π‘₯π‘₯ π‘₯π‘₯π‘₯π‘₯π‘₯π‘₯ πŸπŸ– π‘₯ π‘₯ π‘₯ 𝟏 πŸ’ 𝒙 5555 𝟏 πŸ’ 𝒙 𝟏 πŸ’ 𝒙 𝟏 πŸ’ 𝒙
Two Step Equations
2 π‘₯+3=23 What is the value of ? πŸπŸ‘ 𝒙 πŸ‘ 𝒙 𝟐𝟎 𝒙 𝒙 𝟏𝟎 𝒙 π‘₯ =10 βˆ’ 3 Γ· 2
3 𝑦 +6=2 7 What is the value of ? 𝑦 =7 πŸπŸ• π’š πŸ” π’š π’š 𝟐𝟏 π’š π’š π’š πŸ• π’š βˆ’ 6 Γ· 3
2a Solving Equations Bar Method.pkdkdmptx
2a Solving Equations Bar Method.pkdkdmptx
Two Step Equations (and balancing)
3 π‘š+10=34 What is the value of ? π‘š =8 βˆ’10 Γ· 3 πŸ‘πŸ’ 3 π‘šπŸπŸŽ πŸπŸ’ 3 π‘š πŸ– π’Ž 3π‘š+10=34 3 π‘š=24 βˆ’10 βˆ’10 Γ· 3 Γ· 3 π‘š=8
4 π‘š+2=20 What is the value of ? π‘š=4.5 βˆ’2 Γ· 2 𝟐𝟎 4 π‘šπŸ πŸπŸ– 4 π‘š πŸ’ . πŸ“ π’Ž 4π‘š+2=20 4 π‘š=18 βˆ’2 βˆ’2 Γ· 4 Γ· 4 π‘š=4.5
2a Solving Equations Bar Method.pkdkdmptx
2 𝑧 βˆ’ 4=6 What is the value of ? 𝑧 =5 + 4 Γ· 2 πŸ” πŸ’ πŸ”πŸ’ πŸ“ 𝒛 2 𝑧 βˆ’ 4=6 2 𝑧=10 +4 +4 Γ· 2 Γ· 2 𝑧 =5
2a Solving Equations Bar Method.pkdkdmptx
𝑔 5 + 2 =6 What is the value of ? 𝑔=20 βˆ’2 Γ—5 𝑔 5 +2 =6 𝑔 5 = 4 βˆ’2 βˆ’2 Γ—5 Γ—5 𝑔=20 𝟐 𝟏 πŸ“ π’ˆ 6 𝟏 πŸ“ π’ˆ 4 4 4 4 𝟏 πŸ“ π’ˆ 𝟏 πŸ“ π’ˆ 𝟏 πŸ“ π’ˆ 4 𝟏 πŸ“ π’ˆ 𝟏 πŸ“ π’ˆ 4
Equations with unknowns on both sides
3 π‘₯+7=π‘₯+13 What is the value of ? π‘₯ =3 βˆ’ π‘₯ βˆ’ 7 πŸπŸ‘ 𝒙 πŸ• 𝒙 𝒙 𝒙 πŸπŸ‘ πŸ• 𝒙 𝒙 πŸ” 𝒙 𝒙 πŸ‘ 𝒙 Γ· 2
4 π‘Ž+5=2 π‘Ž+23 What is the value of ? π‘Ž= βˆ’2 π‘Ž βˆ’ 5 πŸπŸ‘ 𝒂 πŸ“ 𝒂𝒂 𝒂𝒂 𝒂 πŸπŸ‘ πŸ“ 𝒂𝒂 πŸπŸ– 𝒂𝒂 πŸ— 𝒂 Γ· 2
4 π‘Ž+5=2 π‘Ž+23 What is the value of ? π‘Ž= βˆ’2 π‘Ž βˆ’ 5 πŸπŸ‘ 𝒂 πŸ“ 𝒂𝒂 𝒂𝒂 𝒂 πŸπŸ‘ πŸ“ 𝒂𝒂 πŸπŸ– 𝒂𝒂 πŸ— 𝒂 Γ· 2
2a Solving Equations Bar Method.pkdkdmptx
3 π‘Ž+2=π‘Ž+12 What is the value of ? π‘Ž=5 βˆ’π‘Ž βˆ’2 𝟏𝟐 𝒂 𝟐 𝒂𝒂 𝒂 Γ· 2 𝟏𝟐 𝟐 𝒂𝒂 𝟏𝟎 𝒂𝒂 πŸ“ 𝒂 3 π‘Ž+2=π‘Ž+12 2 π‘Ž+2=12 βˆ’ π‘Ž βˆ’2 βˆ’2 βˆ’2 2 π‘Ž=10 Γ· 2 Γ· 2 π‘Ž=5
5 π‘Ž+6=2 π‘Ž+36 What is the value of ? π‘Ž=10 βˆ’2π‘Ž βˆ’6 Γ· 3 5π‘Ž+6=2π‘Ž+36 3 π‘Ž+6=36 βˆ’2π‘Ž βˆ’2π‘Ž βˆ’6 βˆ’6 3 π‘Ž=30 Γ· 3 Γ· 3 π‘Ž=10 πŸ‘πŸ” 𝒂 πŸ” 𝒂𝒂 𝒂𝒂 𝒂𝒂 πŸ‘πŸ” πŸ” 𝒂𝒂𝒂 πŸ‘πŸŽ 𝒂𝒂𝒂 1 𝒂
2a Solving Equations Bar Method.pkdkdmptx
4 π‘‘βˆ’11=2𝑑+15 What is the value of ? 𝑑=13 βˆ’2 𝑑 Γ· 2 πŸπŸ“ 𝟏𝟏 4 𝒅 𝒅 πŸπŸ“ 𝟏𝟏 2 πŸπŸ“πŸπŸ 𝒅 𝒅 πŸπŸ‘ 𝒅 +11 4 π‘‘βˆ’11=2𝑑+15 2 𝑑 βˆ’11=15 βˆ’2 𝑑 βˆ’2 𝑑 +11 +11 2 𝑑=26 Γ· 2 Γ· 2 𝑑=13
2a Solving Equations Bar Method.pkdkdmptx

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2a Solving Equations Bar Method.pkdkdmptx

  • 1. Monday 3 March 2025 Solving Equations with the Bar Method Starte r C W
  • 2. VALUE OF A FUNCTION Monday 3 March 202 5 Solving Equations with the Bar Method
  • 5. What is the value of ? π‘Ž βˆ’ 25=35 35 25
  • 6. π‘₯+25=35 10=π‘₯+8 π‘₯ βˆ’6=18 10=π‘₯ βˆ’2 80 +π‘₯=60 10=20 βˆ’ π‘₯ Can you represent these using a bar method. What is the value of ? 35 25 10 8 60 80 18 6 10 2 10
  • 7. 2 𝑣=2 8 What is the value of ? 28
  • 8. π‘š Γ· 4=6 What is the value of ? π’Ž 6 6 6 6
  • 9. 5 π‘₯=35 80=10 π‘₯ 18 Γ· π‘₯=3 π‘₯ Γ· 4=5 Can you represent these using a bar method. What is the value of ? 35 π‘₯ π‘₯ π‘₯ π‘₯ π‘₯ 80 π‘₯π‘₯π‘₯π‘₯ π‘₯π‘₯π‘₯π‘₯π‘₯π‘₯ πŸπŸ– π‘₯ π‘₯ π‘₯ 𝟏 πŸ’ 𝒙 5555 𝟏 πŸ’ 𝒙 𝟏 πŸ’ 𝒙 𝟏 πŸ’ 𝒙
  • 11. 2 π‘₯+3=23 What is the value of ? πŸπŸ‘ 𝒙 πŸ‘ 𝒙 𝟐𝟎 𝒙 𝒙 𝟏𝟎 𝒙 π‘₯ =10 βˆ’ 3 Γ· 2
  • 12. 3 𝑦 +6=2 7 What is the value of ? 𝑦 =7 πŸπŸ• π’š πŸ” π’š π’š 𝟐𝟏 π’š π’š π’š πŸ• π’š βˆ’ 6 Γ· 3
  • 15. Two Step Equations (and balancing)
  • 16. 3 π‘š+10=34 What is the value of ? π‘š =8 βˆ’10 Γ· 3 πŸ‘πŸ’ 3 π‘šπŸπŸŽ πŸπŸ’ 3 π‘š πŸ– π’Ž 3π‘š+10=34 3 π‘š=24 βˆ’10 βˆ’10 Γ· 3 Γ· 3 π‘š=8
  • 17. 4 π‘š+2=20 What is the value of ? π‘š=4.5 βˆ’2 Γ· 2 𝟐𝟎 4 π‘šπŸ πŸπŸ– 4 π‘š πŸ’ . πŸ“ π’Ž 4π‘š+2=20 4 π‘š=18 βˆ’2 βˆ’2 Γ· 4 Γ· 4 π‘š=4.5
  • 19. 2 𝑧 βˆ’ 4=6 What is the value of ? 𝑧 =5 + 4 Γ· 2 πŸ” πŸ’ πŸ”πŸ’ πŸ“ 𝒛 2 𝑧 βˆ’ 4=6 2 𝑧=10 +4 +4 Γ· 2 Γ· 2 𝑧 =5
  • 21. 𝑔 5 + 2 =6 What is the value of ? 𝑔=20 βˆ’2 Γ—5 𝑔 5 +2 =6 𝑔 5 = 4 βˆ’2 βˆ’2 Γ—5 Γ—5 𝑔=20 𝟐 𝟏 πŸ“ π’ˆ 6 𝟏 πŸ“ π’ˆ 4 4 4 4 𝟏 πŸ“ π’ˆ 𝟏 πŸ“ π’ˆ 𝟏 πŸ“ π’ˆ 4 𝟏 πŸ“ π’ˆ 𝟏 πŸ“ π’ˆ 4
  • 22. Equations with unknowns on both sides
  • 23. 3 π‘₯+7=π‘₯+13 What is the value of ? π‘₯ =3 βˆ’ π‘₯ βˆ’ 7 πŸπŸ‘ 𝒙 πŸ• 𝒙 𝒙 𝒙 πŸπŸ‘ πŸ• 𝒙 𝒙 πŸ” 𝒙 𝒙 πŸ‘ 𝒙 Γ· 2
  • 24. 4 π‘Ž+5=2 π‘Ž+23 What is the value of ? π‘Ž= βˆ’2 π‘Ž βˆ’ 5 πŸπŸ‘ 𝒂 πŸ“ 𝒂𝒂 𝒂𝒂 𝒂 πŸπŸ‘ πŸ“ 𝒂𝒂 πŸπŸ– 𝒂𝒂 πŸ— 𝒂 Γ· 2
  • 25. 4 π‘Ž+5=2 π‘Ž+23 What is the value of ? π‘Ž= βˆ’2 π‘Ž βˆ’ 5 πŸπŸ‘ 𝒂 πŸ“ 𝒂𝒂 𝒂𝒂 𝒂 πŸπŸ‘ πŸ“ 𝒂𝒂 πŸπŸ– 𝒂𝒂 πŸ— 𝒂 Γ· 2
  • 27. 3 π‘Ž+2=π‘Ž+12 What is the value of ? π‘Ž=5 βˆ’π‘Ž βˆ’2 𝟏𝟐 𝒂 𝟐 𝒂𝒂 𝒂 Γ· 2 𝟏𝟐 𝟐 𝒂𝒂 𝟏𝟎 𝒂𝒂 πŸ“ 𝒂 3 π‘Ž+2=π‘Ž+12 2 π‘Ž+2=12 βˆ’ π‘Ž βˆ’2 βˆ’2 βˆ’2 2 π‘Ž=10 Γ· 2 Γ· 2 π‘Ž=5
  • 28. 5 π‘Ž+6=2 π‘Ž+36 What is the value of ? π‘Ž=10 βˆ’2π‘Ž βˆ’6 Γ· 3 5π‘Ž+6=2π‘Ž+36 3 π‘Ž+6=36 βˆ’2π‘Ž βˆ’2π‘Ž βˆ’6 βˆ’6 3 π‘Ž=30 Γ· 3 Γ· 3 π‘Ž=10 πŸ‘πŸ” 𝒂 πŸ” 𝒂𝒂 𝒂𝒂 𝒂𝒂 πŸ‘πŸ” πŸ” 𝒂𝒂𝒂 πŸ‘πŸŽ 𝒂𝒂𝒂 1 𝒂
  • 30. 4 π‘‘βˆ’11=2𝑑+15 What is the value of ? 𝑑=13 βˆ’2 𝑑 Γ· 2 πŸπŸ“ 𝟏𝟏 4 𝒅 𝒅 πŸπŸ“ 𝟏𝟏 2 πŸπŸ“πŸπŸ 𝒅 𝒅 πŸπŸ‘ 𝒅 +11 4 π‘‘βˆ’11=2𝑑+15 2 𝑑 βˆ’11=15 βˆ’2 𝑑 βˆ’2 𝑑 +11 +11 2 𝑑=26 Γ· 2 Γ· 2 𝑑=13