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Solve the system 3x + 2y = 16 and 5x - 2y = 0

Worked through step by step. More systems.

Answer
x=2,y=5\boxed{x = 2, y = 5}

Substituting x=2x = 2 and y=5y = 5 into the original equations gives 3(2)+2(5)=163(2) + 2(5) = 16 and 5(2)−2(5)=05(2) - 2(5) = 0, which are both true.

Problem: Solve the system of linear equations 3x+2y=163x + 2y = 16 and 5x−2y=05x - 2y = 0.

Steps

  1. Add the two equations
    (3x+2y)+(5x−2y)=16+0(3x + 2y) + (5x - 2y) = 16 + 0
  2. Simplify the equation
    8x=168x = 16
  3. Divide both sides by 8
    x=2x = 2
  4. Substitute x=2x = 2 into the second equation
    5(2)−2y=05(2) - 2y = 0
  5. Simplify the equation
    10−2y=010 - 2y = 0
  6. Subtract 10 from both sides
    −2y=−10-2y = -10
  7. Divide both sides by -2
    y=5y = 5

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