Solving Systems of Equations.

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Solving Systems of three equations with three variables Using substitution or elimination.

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Presentation transcript:

Solving Systems of Equations. Elimination Method Solving Systems of Equations.

4x + 3y = 10 2x + y = 4 Elimination Substitution 1 2 Pick one of the variables you would eliminate. Find the LCM . Multiply equations to get the LCM and make sure they are opposite integers. Eliminate by adding. Look for a variable with a coefficient Of 1. Isolate that variable. 2x + y = 4 y = -2x + 4 Now substitute y with -2x +4 into the first equation. Want to eliminate y variable. LCM = 3 4x + 3y = 10 Solve for y: 4 (1) + 3y = 10 4 + 3y = 10 -4 - 4 3y = 6 y = 2 2x + y = 4 -3( 2x + y = 4) = -6x – 3y -12 4x + 3 (-2x + 4) = 10 4x – 6x + 12 = 10 -2x + 12 = 10 - 12 - 12 -2x = -2 X = 1 Solve for y: 4 (1) + 3y = 10 4 + 3y = 10 -4 - 4 3y = 6 y = 2 4x + 3y = 10 -6x – 3y = -12 -2x = -2 X = 1

Which one would you use? Substitution or Elimination or Both? 5t + 2d = 80 -5t – 20d = -350 2x + y = 3 4x + 3y = 5 6x + 7y = 10 2x - 3y = 14 X + 3y = -2 -3x + 5y = 6

Consistent, Inconsistent Systems Dependent, Independent Equations (No Solution) (1 or more solutions) Consistent System Inconsistent System Dependent Equation All solution Independent Equation 1 or no solutions