SYSTEMS OF LINEAR EQUATIONS Solving Linear Systems Algebraically

Solving Systems of Equations Algebraically When you graph, sometimes you cannot find the exact point of intersection. We can use algebra to find the exact point. 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 Algebraically We will look at TWO methods to solve systems algebraically: 1) Substitution 2) Elimination

Method 1: Substitution Steps: Choose one of the two equations and isolate one of the variables. Substitute the new expression into the other equation for the variable. Solve for the remaining variable. Substitute the solution into the other equation to get the solution to the second variable.

Method 1: Substitution Example: Equation ‘a’: 3x + 4y = - 4 Equation ‘b’: x + 2y = 2 Isolate the ‘x’ in equation ‘b’: x = - 2y + 2

Method 1: Substitution 3(- 2y + 2) + 4y = - 4 Example, continued: Equation ‘a’: 3x + 4y = - 4 Equation ‘b’: x + 2y = 2 Substitute the new expression, x = - 2y + 2 for x into equation ‘a’: 3(- 2y + 2) + 4y = - 4

Method 1: Substitution 3(- 2y + 2) + 4y = - 4 - 6y + 6 + 4y = - 4 Example, continued: Equation ‘a’: 3x + 4y = - 4 Equation ‘b’: x + 2y = 2 Solve the new equation: 3(- 2y + 2) + 4y = - 4 - 6y + 6 + 4y = - 4 - 2y + 6 = - 4 - 2y = - 10 y = 5

Method 1: Substitution x + 2 (5) = 2 x + 10 = 2 x = - 8 Example, continued: Equation ‘a’: 3x + 4y = - 4 Equation ‘b’: x + 2y = 2 Substitute y = 5 into either equation ‘a’ or ‘b’: x + 2 (5) = 2 x + 10 = 2 x = - 8 The solution is (-8, 5).

Method 2: Elimination Steps: Line up the two equations using standard form (Ax + By = C). GOAL: The coefficients of the same variable in both equations should have the same value but opposite signs. 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: Elimination Add the two equations (like terms). Steps, continued: Add the two equations (like terms). The variable with opposite coefficients should be eliminated. Solve for the remaining variable. 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: Elimination Add the equations (the x’s are eliminated): Example, continued: Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26 Equation ‘b’: 4x - 5y = 8 ------> 4x - 5y = 8 Add the equations (the x’s are eliminated): -4x + 8y = -26 4x - 5y = 8 3y = -18 y = -6

Method 2: Elimination Solution: ( , -6) Example, continued: Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26 Equation ‘b’: 4x - 5y = 8 ------> 4x - 5y = 8 Substitute y = -6 into either equation: 4x - 5(-6) = 8 4x + 30 = 8 4x = -22 x = Solution: ( , -6)

Method 2: Elimination Equation ‘a’: -9x + 6y = 0 Example 2: Equation ‘a’: -9x + 6y = 0 Equation ‘b’: -12x + 8y = 0 Multiply equation ‘a’ by –4 and equation ‘b’ by 3 to eliminate the x’s: Equation ‘a’: - 4(-9x + 6y = 0) Equation ‘b’: 3(-12x + 8y = 0)

Method 2: Elimination 36x - 24y = 0 -36x + 24y = 0 0 = 0 Example 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 0 = 0 Example 2, continued: When 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.

Method 2: Elimination 36x - 24y = 0 -36x + 24y = 0 0 = 0 Example 2, continued: Since 0 = 0 is TRUE, there are infinite solutions.

Solving Systems of Three Equations Algebraically When we have three equations in a system, we can use the same two methods to solve them algebraically as with two equations. Whether you use substitution or elimination, you should begin by numbering the equations!

Solving Systems of Three Equations Substitution Method Choose one of the three equations and isolate one of the variables. Substitute the new expression into each of the other two equations. These two equations now have the same two variables. Solve this 2 x 2 system as before. Find the third variable by substituting the two known values into any equation.

Solving Systems of Three Equations Linear Combination Method Choose two of the equations and eliminate one variable as before. Now choose one of the equations from step 1 and the other equation you didn’t use and eliminate the same variable. You should now have two equations (one from step 1 and one from step 2) that you can solve by elimination. Find the third variable by substituting the two known values into any equation.