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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 1 1 Chapter 4 Systems of Linear Equations and Inequalities.

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Presentation on theme: "Copyright © 2015, 2011, 2007 Pearson Education, Inc. 1 1 Chapter 4 Systems of Linear Equations and Inequalities."— Presentation transcript:

1 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 1 1 Chapter 4 Systems of Linear Equations and Inequalities

2 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 2 Systems of Equations and Inequalities 4.1Solving Systems of Linear Equations in Two Variables 4.2Solving Systems of Linear Equations in Three Variables 4.3Solving Applications Using Systems of CHAPTER 4

3 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 3 Solving Systems of Linear Equations in Two Variables 1.Determine whether an ordered pair is a solution for a system of equations. 2.Solve systems of linear equations graphically and classify systems. 3.Solve systems of linear equations using substitution. 4.Solve systems of linear equations using elimination. 4.1

4 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 4 System of equations: A group of two or more equations. Solution for a system of equations: An ordered set of numbers that makes all equations in the system true.

5 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 5 Checking a Solution to a System of Equations To verify or check a solution to a system of equations, 1. Replace each variable in each equation with its corresponding value. 2. Verify that each equation is true.

6 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 6 Example Determine whether each ordered pair is a solution to the system of equations. a. (  3, 2)b. (3, 4) Solution a. (  3, 2) x + y = 7y = 3x – 2  3 + 2 = 7 2 = 3(  3) – 2  1 = 72 =  11 FalseFalse Because (−3, 2) does not satisfy both equations, it is not a solution for the system.

7 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 7 continued b. (3, 4) x + y = 7y = 3x – 2 3 + 4 = 7 4 = 3(3) – 2 7 = 74 = 7 TrueFalse Because (3, 4) satisfies only one equation, it is not a solution to the system of equations.

8 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 8 A system of two linear equations in two variables can have one solution, no solution, or an infinite number of solutions. The lines intersect at a single point. There is one solution. The equations have the same slope, the lines are parallel. There is no solution. The lines are identical. There are an infinite number of solutions.

9 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 9 Solving Systems of Equations Graphically To solve a system of linear equations graphically, 1. Graph each equation. a. If the lines intersect at a single point, then the coordinates of that point form the solution. b. If the lines are parallel, there is no solution. c. If the lines are identical, there are an infinite number of solutions, which are the coordinates of all the points on that line. 2. Check your solution.

10 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 10 Example Solve the system of equations graphically. Solution Graph each equation. The lines intersect at a single point, (  2, 4). We can check the point in each equation to verify and we leave that to you. y = 2 – x 2x + 4y = 12 (  2, 4)

11 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 11 Example Solve the system of equations graphically. Solution Graph each equation. The lines appear to be parallel, which we can verify by comparing the slopes. The slopes are the same, so the lines are parallel. The system has no solution.

12 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 12 Example Solve the system of equations graphically. Solution Graph each equation. The lines appear to be identical. The equations are identical. All ordered pairs along the line are solutions.

13 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 13 Consistent system of equations: A system of equations that has at least one solution. Inconsistent system of equations: A system of equations that has no solution. Dependent linear equations in two unknowns: Equations with identical graphs. Independent linear equations in two unknowns: Equations that have different graphs.

14 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 14 Classifying Systems of Equations To classify a system of two linear equations in two unknowns, write the equations in slope-intercept form and compare the slopes and y-intercepts. Consistent system with independent equations: The system has a single solution at the point of intersection. The graphs are different. They have different slopes. Consistent system with dependent equations: The system has an infinite number of solutions. The graphs are identical. They have the same slope and same y-intercept. Inconsistent system: The system has no solution. The graphs are parallel lines. They have the same slope, but different y- intercepts.

15 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 15 Example For each of the systems of equations in the example, determine whether the system is consistent with independent equations, consistent with dependent equations, or inconsistent. How many solutions does the system have? 2a. The graphs intersected at a single point. The system is consistent. The equations are independent (different graphs), and the system has one solution: (  2, 4).

16 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 16 Example For each of the systems of equations in the example, determine whether the system is consistent with independent equations, consistent with dependent equations, or inconsistent. How many solutions does the system have? 2b. The graphs were parallel lines. The system is inconsistent and has no solutions.

17 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 17 Example For each of the systems of equations in the example, determine whether the system is consistent with independent equations, consistent with dependent equations, or inconsistent. How many solutions does the system have? 2c. The graphs coincide. The system is consistent (it has a solution) with dependent equations (same graph) and has an infinite number of solutions.

18 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 18 Solving Systems of Two Linear Equations Using Substitution To find the solution of a system of two linear equations using the substitution method, 1. If necessary, isolate one of the variables in one of the equations. 2. In the other equation, substitute the expression you found in step 1 for that variable. 3. Solve this new equation. (It now has only one variable.) 4. Using one of the equations containing both variables, substitute the value you found in step 3 for that variable and solve for the value of the other variable. 5. Check the solution in the original equations.

19 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 19 Example Solve the system of equations using substitution. Solution Step 1: Isolate a variable in one equation. The second equation is solved for x. Step 2: Substitute x = 1 – y for x in the first equation.

20 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 20 continued Step 3: Solve for y. Step 4: Solve for x by substituting 4 for y in one of the original equations. The solution is (  3, 4).

21 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 21 Example Solve the system of equations using substitution. Solution Step 1: Isolate a variable in one equation. Use either equation. 2x + y = 7 y = 7 – 2x Step 2: Substitute y = 7 – 2x for y in the second equation.

22 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 22 continued Step 3: Solve for x. Step 4: Solve for y by substituting 2 for x in one of the original equations. The solution is (2, 3).

23 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 23 Example Solve the system using elimination. Solution We can add the equations. Notice that y is eliminated, so we can easily solve for the value of x. Divide both sides by 4 to isolate x. Now that we have the value of x, we can find y by substituting 4 for x in one of the original equations. x + y = 9 4 + y = 9 y = 5 The solution is (4, 5).

24 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 24 Example Solve the system of equations. Solution Because no variables are eliminated if we add, we rewrite one of the equations so that it has a term that is the additive inverse of one of the terms in the other equation. Multiply the each term of the first equation by 4. Solve for y. x + y = 8 3 + y = 8 y = 5 The solution is (3, 5).

25 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 25 Example Solve the system of equations. Solution Choose either variable to eliminate. We can multiply both equations by numbers that make the terms additive inverses to cancel them. Multiply the first equation by 2. Multiply the second equation by  5. Multiply by 2. Multiply by  5. Add the equations to eliminate y.

26 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 26 continued Substitute x = 1 into either of the original equations. 4x – 5y = 19 4(1) – 5y = 19 4 – 5y = 19 –5y = 15 y =  3 The solution is (1, –3).

27 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 27 Example Solve the system of equations. Solution To clear the decimals in Equation 1, multiply by 100. To clear the fractions in Equation 2, multiply by 5. Multiply by 100. Multiply by 5. Multiply equation 2 by  1 then combine the equations.

28 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 28 continued Substitute to find y. The solution is (  1,  3).

29 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 29 Solving Systems of Two Linear Equations Using Elimination To solve a system of two linear equations using the elimination method, 1. Write the equations in standard form (Ax + By = C). 2. Use the multiplication principle to clear fractions or decimals (optional). 3. If necessary, multiply one or both equations by a number (or numbers) so that they have a pair of terms that are additive inverses. 4. Add the equations. The result is an equation in terms of one variable. 5. Solve the equation from step 4 for the value of that variable. 6. Using an equation containing both variables, substitute the value you found in step 5 for the corresponding variable and solve for the value of the other variable. 7. Check your solution in the original equations.

30 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 30 Inconsistent Systems and Dependent Equations When both variables have been eliminated and the resulting equation is false, such as 0 = 5, there is no solution. The system is inconsistent. When both variables have been eliminated and the resulting equation is true, such as 0 = 0, the equations are dependent. There are an infinite number of solutions.

31 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 31 Example Solve the system of equations. Solution Notice that the left side of the equations are additive inverses. Adding the equations will eliminate both variables. False statement. The system is inconsistent and has no solution.

32 Copyright © 2015, 2011, 2007 Pearson Education, Inc. 32 Example Solve the system of equations using substitution. Solution Substitute y = 4 – 3x into the second equation. True statement. The number of solutions is infinite.


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