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Solving Equations with Variables on Both Sides

  • To solve equations with variables on both sides, bring all the variable terms to one side of the equation. Use the Distributive Property, if necessary, collect like terms, and simplify.
  • To solve equations with variables on both sides, simplify as much as you can on both sides before you apply inverse operations to isolate the variable.
  • After finding the solution, substitute to check your answer. In real life situations, ask yourself if the answer is reasonable.

Solving Equations with Variables on Both Sides

5x + 15 = 8x. Solve for x.

  • 5x − 8x = − 15
  • − 3x = − 15

x = 3

− 8 − 4x = 10 + 3x. Solve for x.

  • − 8 − 10 = 3x + 7x
  • − 18 = 10x
  • x = − [18/10]

x = − [9/5]

4x + 5 = 3(2x + 1). Solve for x.

  • 4x + 5 = 6x + 3
  • 4x − 6x = 3 − 5
  • − 2x = − 2

x = 1

11 − 5x + 3 = − 3x − 10. Solve for x.

  • − 5x + 3x = − 10 − 11 − 3
  • − 2x = − 24

x = 12

3.6x = 4(x − 2). Solve for x.

  • 3.6x = 4x − 8
  • 3.6x − 4x = − 84
  • − 0.4x = − 8

x = 20

35 + 7x = 24 + x + 1. Solve for x.

  • 7x − x = 24 + 1 − 35
  • 6x = 25 − 35
  • 6x = − 104

x = − [5/3]

14.5 + 5x = − 3.3x − 10.4. Solve for x.

  • 5x + 3.3x = − 10.4 − 14.5
  • 8.3x = − 24.9

x = − 3

− 3(5x − 2) = − 18x + 10x. Solve for x.

  • − 15x + 6 = − 8x
  • − 15x + 8x = − 6
  • − 7x = − 6

x = [6/7]

Two boats are taking the same path to a destination. Ship A travels at 25 mi/hr, while ship B travels at 30 mi/hr and leaves 30 min after ship A does. How long after ship A leaves will ship B catch up to ship A?

  • Let t = time that ship A travels before ship B catches up.
  • distanceship A = distanceship B
    distance = rate × time
  • 25 mi/hr ×t = 30 mi/hr × (t - 0.5)
  • 25t = 30t − 15
  • 25t − 30t = − 15
  • − 5t = − 15

t = 3 hours

Lisa runs [1/8] mi/min. George runs [1/10] mi/min. Lisa starts running 5 min after George starts. How long after George starts running have they run the same distance?

  • Let t = number of minutes
  • distanceLisa = distanceGeorge
    distance = rate × time
  • [1/8](t − 5) = [1/10]t
  • [1/8]t − [5/8] = [1/10]t
  • [1/8]t − [1/10]t = [5/8]
  • [5/40]t − [4/40]t = [25/40]
  • [1/40]t = [25/40]

t = 25 min

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Solving Equations with Variables on Both Sides

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Mathematics: Pre Algebra