Chapter 4.1 Solving Systems of Linear Equations in two variables.

Presentation on theme: "Chapter 4.1 Solving Systems of Linear Equations in two variables."— Presentation transcript:

1 Chapter 4.1 Solving Systems of Linear Equations in two variables.

2 Objectives Solve a system by graphing Solve a system by substitution Solve a system by elimination

3 Solution of a system A solution of a system of two equations in two variables is an ordered pair (x,y) that makes both equations true.

4 Example 1: Determine whether the given ordered pair is a solution of the system.

5 Example 1a: Continued Replace x with -1 and y with 1 in each equation. -x + y = 22x – y = -3 -(-1) + (1) = 22(-1) – (1) = -3 1 +1 = 2-2 – 1 = -3 2 = 2 -3 = -3 True *Since (-1,1) makes both equations true, it is a solution. Using set notation, the solution set is {(-1,1)}

6 Example 1b: Continued We replace x with -2 and y with 3 in each equation 5x + 3y = -1x – y = 1 5(-2) + 3(3) = -1(-2) – (3) = 1 -10 + 9 = -1 -5 = 1 -1 = -1 -5 = 1 True False **Since the ordered pair (-2,3) does not make BOTH equations true, it is not a solution of the system.

7 Helpful Hint Reading values from graphs may not be accurate. Until a proposed solution is checked in both equations of the system, we can only assume that we have estimated a solution.

8 Example 2: Solve each system by graphing Since the graph of a linear equation in two variables is a line, graphing two such equations yields two lines in a plane.

9 Step 1: Graph both lines y = -x + 2 y – int: (0,2) m = -1 -y = -3x – 2 y = 3x + 2 y – int: (0,2) m = 3 (0,2) Common Solution

10 Step 2: Substitute in common solution to determine if ordered pair satisfies both equations. *(0,2) does satisfy both equations. We conclude therefore that (0,2) is the solution of the system. A system that has at least one solution, such as this one, is said to be consistent.

11 Example 2: B Step 1: Start by graphing each line.

12 Graph each line If parallel, system has no solution To check that lines are parallel, write each equation in point slope form. A system that has not solution is said to be inconsistent.

13 Example 2: C The graph of each equation appears to be in the same line. This means that the equations have identical solutions. Any ordered pair solution of one equation satisfies the other equation also. Thus these equations are said to be dependent equations. The solution set is {(x,y)/ x + 2y = 5}

14 Concept Check The equations in the system are dependent and the system has an infinite number of solutions. Which ordered pairs below are solutions? a. (4,0)b. (-4,0)c. (-1, 1)

15 Summary of Solutions One Solution: Consistent System Independent equations No Solution: Inconsistent System Independent equations Infinite number of solutions: Consistent system Dependent equations

16 Concept Check How can you tell just by looking at the following system that it has no solution?

17 Concept Check How can you tell just by looking at the following system that it has infinitely many solutions?

18 Example 3: Solving equations by substitution method Use the substitution method to solve the system.

19 Example 3: Continued Substitute 2y – 5 for x in the first equation. 2 (2y – 5) + 4y = -6 4y – 10 + 4y = -6 8 y – 10 = -6 8 y = 4 y = ½ The y – coordinate of the solution is 1/2. To find the x – coordinate, we replace y with ½ in the second equation.

20 Example 3: Continued x = 2y – 5 x = 2(1/2) – 5 x = 1 – 5 x = -4 The ordered pair solution is (-4, ½ ). Check to see that (-4,1/2) satisfies both equations of the system.

21 Solving a system of two equations using the substitution method Step 1: Solve one of the equations for one of its variables. Step 2: Substitute the expression for the variable found in step 1 into the other equation. Step 3: Find the value of one variable by solving the equation from step 2. Step 4: Find the value of the other variable by substituting the value found in Step 3 into the equation from Step 1. Step 5: Check the ordered pair solution in both original equations.

22 Give it a try! Use the substitution method to solve the system.

23 Example 4 Use the substitution method to solve the system.

24 Example 4 : Continued Step 1: Multiply each equation by its lcm to clear the system of fractions.

25 Example 4: Continued Step 2: Solve the first equation for x - x + 3y = 3 -x = -3y + 3 x = 3y – 3 Step 3: Replace x with 3y – 3 in the second equation. 4(3y -3) – 2y = -9 12 y – 12 – 2y = -9 10 y – 12 = -9 10 y = 3 y =

26 Example 4: Continued Step 4: Replace y within the equation x = 3y – 3 and solve for x Step 5: write your ordered pair solution and check to see that this solution satisfies both equations.

27 Give it a try! Use the substitution method to solve the system:

28 Elimination Method The elimination method or addition method is a second algebraic technique for solving systems of equations. For this method, we rely on a version of the addition property of addition, which states that “equals added to equals are equal.” If A = B and C = D then A+C = B + D

29 Example 5: Use the elimination method to solve the system: Since the left side of each equation is equal to the right side, we add equal quantities by adding the lefts sides of the equations and right sides of the equations.

30 Example 5: Continued x – 5y = -12 -x + y = 4 -4y = -8 y = 2 The y- coordinate is 2. To find the corresponding x- coordinate, we replace y with 2 in either equation. -x + (2) = 4 - x = 2 x = -2 The ordered pair solution is (-2, 2). Check to see that (-2,2) satisfies both equations of the system.

31 Solving a system of two linear equations using the elimination method Step 1: Rewrite each equation in standard form, Ax +By = C. Step 2: If necessary, multiply one or both equations by the same nonzero number so that the coefficient of one variable in one equation is the opposite of its coefficient in the other equation. Step 3: Add the equations. Step 4: Find the value of one variable by solving the equation from step 3. Step 5: Find the value of the second variable by substituting the value found in step 4 into either original equation. Step 6: Check the proposed ordered pair solution in both original equations.

32 Give it a try! Use the elimination method to solve the system:

33 Example 6: Use the elimination method to solve the system. Step 1: Multiply both sides of the first equation by 3 and both sides of the second equation by -2.

34 Example 6: Continued After multiplying you end up with 9x – 6 y = 30 -8x + 6y =-30 x = 0 Substitute 0 for x in either equation to find y – value 3(0) – 2y = 10 -2y = 10 y = -5 Ordered Pair solution: (0, -5) – Check to see that this ordered pair satisfies both equations.

35 Give it a try! Use the elimination method to solve the system:

36 Example 7 Use the elimination method to solve the system. Step 1: Clear the fractions by multiply by sides of 1 st equation by -2. 0 = 1 False No solution. The solution set is { }. This system is inconsistent, and the graphs of the equations are parallel lines.

37 Example 8: Use the elimination method to solve the system.

38 Helpful Hint Remember that not all ordered pairs are solutions of the system in Example 8. Only the infinite number of ordered pairs that satisfy -5x – 3y = 9 or equivalently 10x + 6y = -18.

39 Give it a try! Use the elimination method to solve the system:

40 Give it a try! Use the elimination method to solve the system: