3.2 solving systems algebraically
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3.2 solving systems algebraically

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3.2 solving systems algebraically 3.2 solving systems algebraically Presentation Transcript

  • 3.2 SYSTEMS OF LINEAR EQUATIONSSolving Linear Systems Algebraically
  • Substitution MethodToday’s objective:1. I will use substitution to solve systems of linear equations in two variables.
  • Solving Systems of Equations Algebraically1. When you graph, sometimes you cannot find the exact point of intersection. We can use algebra to find the exact point.2. Also, we do not need to put every equation in slope-intercept form in order to determine if the lines are parallel or the same line. Algebraic methods will give us the same information.
  • Methods of Solving Systems AlgebraicallyWe will look at TWO methods to solvesystems algebraically: 1) Substitution 2) Elimination
  • Method 1: SubstitutionSteps: 1. Choose one of the two equations and isolate one of the variables. 2. Substitute the new expression into the other equation for the variable. 3. Solve for the remaining variable. 4. Substitute the solution into the other equation to get the solution to the second variable.
  • Method 1: SubstitutionExample: Equation ‘a’: 3x + 4y = - 4 Equation ‘b’: x + 2y = 2Isolate the ‘x’ in equation ‘b’: x = - 2y + 2
  • Method 1: SubstitutionExample, continued: Equation ‘a’: 3x + 4y = - 4 Equation ‘b’: x + 2y = 2Substitute the new expression,x = - 2y + 2 for x into equation ‘a’: 3(- 2y + 2) + 4y = - 4
  • Method 1: SubstitutionExample, continued: Equation ‘a’: 3x + 4y = - 4 Equation ‘b’: x + 2y = 2Solve the new equation: 3(- 2y + 2) + 4y = - 4 - 6y + 6 + 4y = - 4 - 2y + 6 = - 4 - 2y = - 10 y= 5
  • Method 1: SubstitutionExample, continued: Equation ‘a’: 3x + 4y = - 4 Equation ‘b’: x + 2y = 2Substitute y = 5 into either equation ‘a’ or ‘b’: x + 2 (5) = 2 x + 10 = 2 x=-8 The solution is (-8, 5).
  • 3.2A Elimination MethodToday’s objective:1. I will use elimination to solve systems of linear equations in two variables.
  • Method 2: EliminationSteps:1. Line up the two equations using standard form (Ax + By = C).2. GOAL: The coefficients of the same variable in both equations should have the same value but opposite signs.3. If this doesn’t exist, multiply one or both of the equations by a number that will make the same variable coefficients opposite values.
  • Method 2: EliminationSteps, continued: 4. Add the two equations (like terms). 5. The variable with opposite coefficients should be eliminated. 6. Solve for the remaining variable. 7. Substitute that solution into either of the two equations to solve for the other variable.
  • Method 2: Elimination Example: Equation ‘a’: 2x - 4y = 13 Equation ‘b’: 4x - 5y = 8 Multiply equation ‘a’ by –2 to eliminate the x’s:Equation ‘a’: -2(2x - 4y = 13)Equation ‘b’: 4x - 5y = 8
  • Method 2: EliminationExample, continued: Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26 Equation ‘b’: 4x - 5y = 8 ------> 4x - 5y = 8Add the equations (the x’s are eliminated): -4x + 8y = -26 4x - 5y = 8 3y = -18 y = -6
  • Method 2: Elimination Example, continued: Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26 Equation ‘b’: 4x - 5y = 8 ------> 4x - 5y = 8Substitute y = -6 into either equation: 4x - 5(-6) = 8 4x + 30 = 8 4x = -22 -22 x= 4 x= 2-11 -11 Solution: ( , -6) 2
  • Method 2: EliminationExample 2: Equation ‘a’: -9x + 6y = 0 Equation ‘b’: -12x + 8y = 0Multiply equation ‘a’ by –4 andequation ‘b’ by 3 to eliminate the x’s: Equation ‘a’: - 4(-9x + 6y = 0) Equation ‘b’: 3(-12x + 8y = 0)
  • Method 2: EliminationExample 2, continued: Equation ‘a’: - 4(-9x + 6y = 0) Equation ‘b’: 3(-12x + 8y = 0) 36x - 24y = 0 -36x + 24y = 0 0=0 What does this answer mean? Is it true?
  • Method 2: Elimination 36x - 24y = 0 Example 2, continued: -36x + 24y = 0 0=0When both variables are eliminated, if the statement is TRUE (like 0 = 0), then they are the same lines and there are infinite solutions. if the statement is FALSE (like 0 = 1), then they are parallel lines and there is no solution.Since 0 = 0 is TRUE, there are infinite solutions.
  • Solving Systems of Three Equations Algebraically1. When we have three equations in a system, we can use the same two methods to solve them algebraically as with two equations.2. Whether you use substitution or elimination, you should begin by numbering the equations!
  • Solving Systems of Three EquationsSubstitution Method 1. Choose one of the three equations and isolate one of the variables. 2. Substitute the new expression into each of the other two equations. 3. These two equations now have the same two variables. Solve this 2 x 2 system as before. 4. Find the third variable by substituting the two known values into any equation.
  • Solving Systems of Three EquationsLinear Combination Method 1. Choose two of the equations and eliminate one variable as before. 2. Now choose one of the equations from step 1 and the other equation you didn’t use and eliminate the same variable. 3. You should now have two equations (one from step 1 and one from step 2) that you can solve by elimination. 4. Find the third variable by substituting the two known values into any equation.
  • -4x – 8y = 16 6x + 12y = -241. Solve the system of linear equations using elimination (linear combination).