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System of Equation

System of Equation. Elimination with multiplication S. Calahan 2008. Multiply one equation to eliminate. 3x + 4y = 6 5x + 2y = - 4 Multiple the second equation by -2 so the coefficients of the y terms are additive inverses. Multiple by -2. 3x + 4y = 6 -2(5x + 2y = - 4) becomes

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System of Equation

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  1. System of Equation Elimination with multiplication S. Calahan 2008

  2. Multiply one equation to eliminate 3x + 4y = 6 5x + 2y = - 4 Multiple the second equation by -2 so the coefficients of the y terms are additive inverses.

  3. Multiple by -2 3x + 4y = 6 -2(5x + 2y = - 4)becomes 3x + 4y = 6 -10x - 4y = 8

  4. Now ADD 3x + 4y = 6 -10x - 4y = 8 -7x = 14 -7 -7 x = -2

  5. Now solve for y 3x + 4y = 6 5x + 2y = - 4 Choose one of the original equations, 3x + 4y = 6, and substitute in X = -2 .

  6. Solve for y 3x + 4y = 6 and x = -2 3(-2) + 4y = 6substitute -2 for x -6 + 4y = 6 solve for y +6 = +6 4y = 12

  7. One Solution 4y = 12 4 4 y = 3 so the solution is (-2, 3)

  8. Multiply both equations to solve. 3x + 4y = -25 2x – 3y = 6

  9. I choose to eliminate the y variable because they already have opposite signs. 3(3x + 4y = -25) 4(2x – 3y = 6) becomes 9x + 12y = -75 8x – 12y = 24

  10. Now add 9x + 12y = -75 8x – 12y = 24 17x = -51 17 17 x = -3

  11. If x = -3 solve for y. 9x + 12y = -75 9(-3) + 12y = -75 -27 + 12y = -75 + 27 +27 12y = -48 12 12 so y = -4

  12. So the solution is (-3, -4)

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