Transcript Part 2
Chapter 3 Systems of Linear Equations Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 1 3.2 Using Substitution and Elimination to Solve Systems Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 2 Example: Solving a System by Substitution Solve the system y x 1 3x 2 y 13 Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 3 Solution From the first equation, we know the value of y is equal to the value of x – 1. So, we substitute x – 1 for y in the second equation: 3x 2 y 13 3 x 2 x 1 13 By making this substitution, we now have an equation in terms of one variable. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 4 Solution Next, solve that equation for x: 3x 2( x 1) 13 3x 2 x 2 13 5x 2 13 5x 15 x3 The x-coordinate of the solution is 3. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 5 Solution To find the y-coordinate, substitute 3 for x in either of the original equations and solve for y: y x 1 y 3 1 y2 So, the solution is (3, 2). Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 6 Solution We can check that (3, 2) satisfies both of the system’s equations: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 7 Solution We can also check that (3, 2) is the solution by graphing the two equations on a graphing calculator and checking that (3, 2) is the intersection point of the two lines. Note, the second equation must be in slope-intercept form first: Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 8 Using Substitution to Solve a Linear System To use substitution to solve a system of two linear equations, 1. Isolate a variable on one side of either equation. 2. Substitute the expression for the variable found in step 1 into the other equation. 3. Solve the equation in one variable found in step 2. 4. Substitute the solution found in step 3 into one of the original equations, and solve for the other variable. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 9 Example: Solving a System of Substitution Solve the system x 3y 4 2x 5 y 5 Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 10 Solution Begin by solving for one of the variables in one of the equations. We can avoid fractions by choosing to solve the first equation for x: x 3y 4 x 3y 4 Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 11 Solution Next, substitute 3y + 4 for x in the second equation and solve for y: 2x 5 y 5 2(3 y 4) 5 y 5 6y 8 5y 5 y 8 5 y 3 Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 12 Solution Finally, substitute –3 for y in the equation x = 3y + 4 and solve for x: x = 3(–3) + 4 x = –5 The solution is (–5, –3). We could verify our work by checking that this ordered pair satisfies both of the original equations. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 13 Adding Left Sides and Adding Right Sides of Two Equations If a = b and c = d, then a+c=b+d In words, the sum of the left sides of two equations is equal to the sum of the right sides. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 14 Example: Solving a System by Elimination Solve the system 4x 5 y 3 3x 5 y 11 Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 15 Solution Begin by adding the left sides and adding the right sides of the two equations: 4x 5 y 3 3x 5 y 11 7 x 0 14 Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 16 Solution Having “eliminated” the variable y, we are left with an equation in one variable. Next, solve that equation for x: 7 x 0 14 7 x 14 x2 Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 17 Solution Substitute 2 for x in either of the original equations and solve for y: 4x 5 y 3 4(2) 5y 3 8 5y 3 5 y 5 y 1 The solution is (2, 1). Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 18 Using Elimination to Solve a Linear System To use elimination to solve a system of two linear equations, 1. If needed, multiply both sides of one equation by a number (and, if necessary, multiply both sides of the other equation by another number) to get the coefficients of one variable to be equal in absolute value and opposite in sign. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 19 Using Elimination to Solve a Linear System 2. Add the left sides and add the right sides of the equations to eliminate one of the variables. 3. Solve the equation in one variable found in step 2. 4. Substitute the solution found in step 3 into one of the original equations, and solve for the other variable. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 20 Example: Solving a System by Elimination Solve the system 4 x 3 y 3 5x 2 y 25 Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 21 Solution 4 x 3 y 3 5x 2 y 25 To eliminate the y terms, multiply both sides of the first equation by 2 and multiply both sides of the second equation by 3. That yields the system 8 x 6 y 6 15x 6 y 75 The coefficients of the y terms are now equal in absolute value and opposite in sign. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 22 Solution Next, add the left sides and add the right sides of the equations and solve for x: 8 x 6 y 6 15 x 6 y 75 23x 0 69 23x 69 x3 Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 23 Solution Substituting 3 for x in the first equation gives 4 x 3 y 3 4(3) 3 y 3 12 3 y 3 3 y 15 y 5 The solution is (3, 5). Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 24 Three Methods for Solving a System Any linear system of two equations can be solved by graphing, substitution, or elimination. All three methods will give the same result. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 25 Example: Using Substitution to Solve an Inconsistent System Consider the linear system y = 2x + 1 y = 2x + 3 The graphs of the equations are parallel lines (why?), so the system is inconsistent and the solution set is the empty set. What happens when we solve this system by substitution? Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 26 Solution We substitute 2x + 1 for y in the second equation and solve for x: We get the false statement 1 = 3. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 27 Inconsistent System of Two Equations If the result of applying substitution or elimination to a linear system of two equations is a false statement, the system is inconsistent – that is, the solution set is the empty set. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 28 Example: Applying Substitution to a Dependent System The system below is dependent and the solution set is the infinite set of solutions of the equation y = 2x + 1. y 2x 1 6 x 3 y 3 What happens when we solve this system by substitution? Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 29 Solution Substitute 2x + 1 for y in the second equation and solve for x: We get the true statement –3 = –3. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 30 Dependent System of Two Linear Equations If the result of applying substitution or elimination to a linear system of two equations is a true statement (one that can be put into the form a = a), then the system is dependent – that is, the solution set is the set of ordered pairs represented by every point on the (same) line. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 31 Using Graphing to Solve an Equation in One Variable To use graphing to solve an equation A = B in one variable, x, where A and B are expressions, 1. Use graphing to solve the system y=A y=B 2. The x-coordinate of any solutions of the system are the solutions of the equation A = B. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 32 Example: Solving an Equation in One Variable by Graphing Solve each equation by referring to the graph at the right. 1 3 1. x 2 x 5 4 2 1 2. x 2 3 4 Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 33 Solution 1. We see the solution of the system 1 y x2 4 3 y x5 2 is the ordered pair (–4, 1). So, the x-coordinate –4 is the solution of the equation 1 3 x 2 x 5 4 2 Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 34 Solution 2. We see the solution of the system 1 y x2 4 y 3 is the ordered pair (4, 3). So, the x-coordinate 4 is the solution of the equation 1 x23 4 Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 35 Example: Solving an Equation in One Variable by Using Tables Use a table to solve 3x – 8 = –7x + 2. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 36 Solution The solution is (1, –5), which has an x-coordinate of 1. So, the solution to the given equation is 1. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 37 Using Tables to Solve an Equation in One Variable To use a table to solve an equation A = B in one variable, x, where A and B are expressions, 1. Use a table to solve the system y=A y=B 2. The x-coordinate of any solutions of the system are the solutions of the equation A = B. Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 3.2, Slide 38