726 CHAPTER 10 Systems of Equations and Inequalities 10.1 Systems of Linear Equations: Sustitution and Elimination PREPARING FOR THIS SECTION Before getting started, review the following: Linear Equations (Appendix, Section A.5, pp. 986 987) Lines (Section 1.4, pp 27 38) Now work the Are You Prepared? prolems on page 737. OBJECTIVES 1 Solve Systems of Equations y Sustitution 2 Solve Systems of Equations y Elimination 3 Identify Inconsistent Systems of Equations Containing Two Variales 4 Express the of a System of Dependent Equations Containing Two Variales 5 Solve Systems of Three Equations Containing Three Variales 6 Identify Inconsistent Systems of Equations Containing Three Variales 7 Express the of a System of Dependent Equations Containing Three Variales We egin with an example. EXAMPLE 1 Movie Theater Ticket Sales A movie theater sells tickets for $8.00 each, with seniors receiving a discount of $2.00. One evening the theater took in $3580 in revenue. If x represents the numer of tickets sold at $8.00 and y the numer of tickets sold at the discounted price of $6.00, write an equation that relates these variales. Each nondiscounted ticket rings in $8.00, so x tickets will ring in 8x dollars. Similarly, y discounted tickets ring in 6y dollars. Since the total rought in is $3580, we must have 8x + 6y = 3580 In Example 1, suppose that we also know that 525 tickets were sold that evening. Then we have another equation relating the variales x and y. The two equations x + y = 525 8x + 6y = 3580 x + y = 525 form a system of equations. In general, a system of equations is a collection of two or more equations, each containing one or more variales. Example 2 gives some illustrations of systems of equations.
SECTION 10.1 Systems of Linear Equations: Sustitution and Elimination 727 EXAMPLE 2 Examples of Systems of Equations (a) 2x + y = 5-4x + 6y = -2 Two equations containing two variales, x and y () x + y2 = 5 2x + y = 4 Two equations containing two variales, x and y (c) x + y + z = 6 c 3x - 2y + 4z = 9 x - y - z = 0 Three equations containing three variales, x, y, and z (d) x + y + z = 5 x - y = 2 Two equations containing three variales, x, y, and z (e) x + y + z = 6 2x + 2z = 4 d y + z = 2 x = 4 Four equations containing three variales, x, y, and z (4) We use a race, as shown, to remind us that we are dealing with a system of equations. We also will find it convenient to numer each equation in the system. A solution of a system of equations consists of values for the variales that are solutions of each equation of the system. To solve a system of equations means to find all solutions of the system. For example, x = 2, y = 1 is a solution of the system in Example 2(a), ecause 2x + y = 5-4x + 6y = -2 2122 + 1 = 4 + 1 = 5-4122 + 6112 = -8 + 6 = -2 A solution of the system in Example 2() is x = 1, y = 2, ecause x + y2 = 5 2x + y = 4 1 + 2 2 = 1 + 4 = 5 2112 + 2 = 2 + 2 = 4 Another solution of the system in Example 2() is x = 11 which you can 4, y = - 3 2, check for yourself. A solution of the system in Example 2(c) is x = 3, y = 2, z = 1, ecause x + y + z = 6 c 3x - 2y + 4z = 9 x - y - z = 0 3 + 2 + 1 = 6 c 3132-2122 + 4112 = 9-4 + 4 = 9 3-2 - 1 = 0 Note that x = 3, y = 3, z = 0 is not a solution of the system in Example 2(c). x + y + z = 6 c 3x - 2y + 4z = 9 x - y - z = 0 3 + 3 + 0 = 6 c 3132-2132 + 4102 = 3 Z 9 3-3 - 0 = 0
728 CHAPTER 10 Systems of Equations and Inequalities Although these values satisfy equations and, they do not satisfy equation. Any solution of the system must satisfy each equation of the system. NOW WORK PROBLEM 9. When a system of equations has at least one solution, it is said to e consistent; otherwise, it is called inconsistent. An equation in n variales is said to e linear if it is equivalent to an equation of the form a 1 x 1 + a 2 x 2 + Á + a n x n = where x 1, x 2, Á, x n are n distinct variales, a 1, a 2, Á, a n, are constants, and at least one of the a s is not 0. Some examples of linear equations are 2x + 3y = 2 5x - 2y + 3z = 10 8x + 8y - 2z + 5w = 0 If each equation in a system of equations is linear, then we have a system of linear equations. The systems in Examples 2(a), (c), (d), and (e) are linear, whereas the system in Example 2() is nonlinear. In this chapter we shall solve linear systems in Sections 10.1 to 10.3. We discuss nonlinear systems in Section 10.6. We egin y discussing a system of two linear equations containing two variales.we can view the prolem of solving such a system as a geometry prolem.the graph of each equation in such a system is a line. So a system of two equations containing two variales represents a pair of lines. The lines either intersect or are parallel or are coincident (that is, identical). 1. If the lines intersect, then the system of equations has one solution, given y the point of intersection. The system is consistent and the equations are independent. See Figure 1(a). 2. If the lines are parallel, then the system of equations has no solution, ecause the lines never intersect. The system is inconsistent. See Figure 1(). 3. If the lines are coincident, then the system of equations has infinitely many solutions, represented y the totality of points on the line. The system is consistent and the equations are dependent. See Figure 1(c). Figure 1 y y y x x x (a) Intersecting lines; system has one solution () Parallel lines; system has no solution (c) Coincident lines; system has infinitely many solutions EXAMPLE 3 Solving a System of Linear Equations Using a Graphing Utility 2x + y = 5-4x + 6y = 12
SECTION 10.1 Systems of Linear Equations: Sustitution and Elimination 729 Figure 2 Y 1 2x 5 10 6 6 2 2 Y 2 3x 2 1 First, we solve each equation for y. This is equivalent to writing each equation in slope intercept form. Equation in slope intercept form is Y 1 = -2x + 5. Equation in slope intercept form is Y 2 = 2 Figure 2 shows the graphs 3 x + 2. using a graphing utility. From the graph in Figure 2, we see that the lines intersect, so the system is consistent and the equations are independent. Using INTERSECT, we otain the solution 11.125, 2.752. Solve Systems of Equations y Sustitution Sometimes to otain exact solutions, we must use algeraic methods. A numer of methods are availale to us for solving systems of linear equations algeraically. In this section, we introduce two methods: sustitution and elimination. We illustrate the method of sustitution y solving the system given in Example 3. EXAMPLE 4 Solving a System of Linear Equations y Sustitution 2x + y = 5-4x + 6y = 12 We solve the first equation for y, otaining 2x + y = 5 y = -2x + 5 Sutract 2x from each side of. We sustitute this result for y into the second equation. The result is an equation containing just the variale x, which we can then solve. -4x + 6y = 12-4x + 61-2x + 52 = 12-4x - 12x + 30 = 12-16x = -18 x = -18-16 = 9 8 Sustitute y = -2x + 5 in. Remove parentheses. Comine like terms and sutract 30 from oth sides. Divide each side y -16. Once we know that x = 9 8, we can easily find the value of y y ack- 9 sustitution, that is, y sustituting use the first equation. 8 for x in one of the original equations. We will 2x + y = 5 y = -2x + 5 y = -2a 9 8 + 5 Sutract 2x from each side. Sustitute x = 9 in. 8 = -9 4 + 20 4 = 11 4 The solution of the system is x = 9 8 = 1.125, y = 11 4 = 2.75.
730 CHAPTER 10 Systems of Equations and Inequalities CHECK: d 2x + y = 5: 2a 9 8 + 11 4 = 9 4 + 11 4 = 20 4 = 5-4x + 6y = 12: -4a 9 8 + 6a 11 4 = - 9 2 + 33 2 = 24 2 = 12 The method used to solve the system in Example 4 is called sustitution. The steps to e used are outlined next. Steps for Solving y Sustitution STEP 1: Pick one of the equations and solve for one of the variales in terms of the remaining variales. STEP 2: Sustitute the result into the remaining equations. STEP 3: If one equation in one variale results, solve this equation. Otherwise repeat Steps 1 and 2 until a single equation with one variale remains. STEP 4: Find the values of the remaining variales y ack-sustitution. STEP 5: Check the solution found. NOW USE SUBSTITUTION TO WORK PROBLEM 19. 2 Solve Systems of Equations y Elimination A second method for solving a system of linear equations is the method of elimination. This method is usually preferred over sustitution if sustitution leads to fractions or if the system contains more than two variales. Elimination also provides the necessary motivation for solving systems using matrices (the suject of Section 10.2). The idea ehind the method of elimination is to replace the original system of equations y an equivalent system so that adding two of the equations eliminates a variale. The rules for otaining equivalent equations are the same as those studied earlier. However, we may also interchange any two equations of the system and/or replace any equation in the system y the sum (or difference) of that equation and a nonzero multiple of any other equation in the system. In Words When using elimination, we want to get the coefficients of one of the variales to e negatives of one another. Rules for Otaining an Equivalent System of Equations 1. Interchange any two equations of the system. 2. Multiply (or divide) each side of an equation y the same nonzero constant. 3. Replace any equation in the system y the sum (or difference) of that equation and a nonzero multiple of any other equation in the system. An example will give you the idea. As you work through the example, pay particular attention to the pattern eing followed. EXAMPLE 5 Solving a System of Linear Equations y Elimination 2x + 3y = 1 -x + y = -3
SECTION 10.1 Systems of Linear Equations: Sustitution and Elimination 731 We multiply each side of equation y 2 so that the coefficients of x in the two equations are negatives of one another. The result is the equivalent system 2x + 3y = 1-2x + 2y = -6 If we add equations and we otain an equation containing just the variale y, which we can solve. 2x + 3y = 1-2x + 2y = -6 5y = -5 y = -1 We ack-sustitute this value for y in equation and simplify. 2x + 3y = 1 2x + 31-12 = 1 2x = 4 x = 2 Sustitute y = -1 in. Simplify. Solve for x. The solution of the original system is x = 2, y = -1. We leave it to you to check the solution. The procedure used in Example 5 is called the method of elimination. Notice the pattern of the solution. First, we eliminated the variale x from the second equation. Then we ack-sustituted; that is, we sustituted the value found for y ack into the first equation to find x. NOW USE ELIMINATION TO WORK PROBLEM 19. Let s return to the movie theater example Example 1. Add and. Solve for y. EXAMPLE 6 Movie Theater Ticket Sales A movie theater sells tickets for $8.00 each, with seniors receiving a discount of $2.00. One evening the theater sold 525 tickets and took in $3580 in revenue. How many of each type of ticket were sold? If x represents the numer of tickets sold at $8.00 and y the numer of tickets sold at the discounted price of $6.00, then the given information results in the system of equations 8x + 6y = 3580 x + y = 525 We use the method of elimination. First, multiply the second equation y -6, and then add the equations. 8x + 6y = 3580-6x - 6y = -3150 2x = 430 x = 215 Add the equations. Since x + y = 525, then y = 525 - x = 525-215 = 310. So 215 nondiscounted tickets and 310 senior discount tickets were sold.
732 CHAPTER 10 Systems of Equations and Inequalities 3 Identify Inconsistent Systems of Equations Containing Two Variales The previous examples dealt with consistent systems of equations that had a single solution. The next two examples deal with two other possiilities that may occur, the first eing a system that has no solution. EXAMPLE 7 An Inconsistent System of Linear Equations 2x + y = 5 4x + 2y = 8 We choose to use the method of sustitution and solve equation for y. 2x + y = 5 y = -2x + 5 Sutract 2x from each side. Now sustitute y = -2x + 5 for y in equation and solve for x. 4x + 2y = 8 Figure 3 10 Y 1 2x 5 4x + 21-2x + 52 = 8 4x - 4x + 10 = 8 0 # x = -2 Sustitute y = -2x + 5 in. Remove parentheses. Sutract 10 from oth sides. This equation has no solution. We conclude that the system itself has no solution and is therefore inconsistent. 6 6 2 Y 2 2x 4 Figure 3 illustrates the pair of lines whose equations form the system in Example 7. Notice that the graphs of the two equations are lines, each with slope -2; one has a y-intercept of 5, the other a y-intercept of 4. The lines are parallel and have no point of intersection. This geometric statement is equivalent to the algeraic statement that the system has no solution. 4 Express the of a System of Dependent Equations Containing Two Variales EXAMPLE 8 Solving a System of Dependent Equations 2x + y = 4-6x - 3y = -12 We choose to use the method of elimination. 2x + y = 4-6x - 3y = -12 6x + 3y = 12-6x - 3y = -12 6x + 3y = 12 0 = 0 Multiply each side of equation y 3. Replace equation y the sum of equations and.
SECTION 10.1 Systems of Linear Equations: Sustitution and Elimination 733 The original system is equivalent to a system containing one equation, so the equations are dependent. This means that any values of x and y for which 6x + 3y = 12 or, equivalently, 2x + y = 4 are solutions. For example, x = 2, y = 0; x = 0, y = 4; x = -2, y = 8; x = 4, y = -4; and so on, are solutions. There are, in fact, infinitely many values of x and y for which 2x + y = 4, so the original system has infinitely many solutions. We will write the solution of the original system either as where x can e any real numer, or as where y can e any real numer. y = -2x + 4 x = - 1 2 y + 2 Figure 4 Y 1 Y 2 2x 4 10 6 6 2 Figure 4 illustrates the situation presented in Example 8. Notice that the graphs of the two equations are lines, each with slope -2 and each with y-intercept 4. The lines are coincident. Notice also that equation in the original system is -3 times equation, indicating that the two equations are dependent. For the system in Example 8, we can write down some of the infinite numer of solutions y assigning values to x and then finding y = -2x + 4. If x = -2, then y = -21-22 + 4 = 8. If x = 0, then y = 4. If x = 2, then y = 0. The ordered pairs 1-2, 82, 10, 42, and 12, 02 are three of the points on the line in Figure 4. NOW WORK PROBLEMS 25 AND 29. 5 Solve Systems of Three Equations Containing Three Variales Just as with a system of two linear equations containing two variales, a system of three linear equations containing three variales also has either exactly one solution (a consistent system with independent equations), or no solution (an inconsistent system), or infinitely many solutions (a consistent system with dependent equations). We can view the prolem of solving a system of three linear equations containing three variales as a geometry prolem.the graph of each equation in such a system is a plane in space. A system of three linear equations containing three variales represents three planes in space. Figure 5 illustrates some of the possiilities. Figure 5 (a) Consistent system; one solution s () Consistent system; infinite numer of solutions (c) Inconsistent system; no solution
734 CHAPTER 10 Systems of Equations and Inequalities Recall that a solution to a system of equations consists of values for the variales that are solutions of each equation of the system. For example, x = 3, y = -1, z = -5 is a solution to the system of equations c x + y + z = -3 2x - 3y + 6z = -21-3x + 5y = -14 3 + 1-12 + 1-52 = -3 2132-31-12 + 61-52 = 6 + 3-30 = -21-3132 + 51-12 = -9-5 = -14 ecause these values of the variales are solutions of each equation. Typically, when solving a system of three linear equations containing three variales, we use the method of elimination. Recall that the idea ehind the method of elimination is to form equivalent equations so that adding two of the equations eliminates a variale. Let s see how elimination works on a system of three equations containing three variales. EXAMPLE 9 Solving a System of Three Linear Equations with Three Variales Use the method of elimination to solve the system of equations. x + y - z = -1 c 4x - 3y + 2z = 16 2x - 2y - 3z = 5 For a system of three equations, we attempt to eliminate one variale at a time, using pairs of equations until an equation with a single variale remains. Our plan of attack on this system will e to use equation to eliminate the variale x from equations and. We egin y multiplying each side of equation y -4 and adding the result to equation. (Do you see why? The coefficients of x are now negatives of one another.) We also multiply equation y -2 and add the result to equation. Notice that these two procedures result in the removal of the variale x from equations and. x y z 1 4x 3y 2z 16 x y z 1 2x 2y 3z 5 Multiply y 4 Multiply y 2 4x 4y 4z 4 4x 3y 2z 16 7y 6z 20 Add. 2x 2y 2z 2 2x 2y 3z 5 4y z 7 Add. x y z 1 7y 6z 20 4y z 7 We now concentrate on equations and, treating them as a system of two equations containing two variales. It is easiest to eliminate z. We multiply each side of equation y 6 and add equations and. The result is the new equation. 7y 6z 20 7y 6z 20 4y z 7 Multiply y 6 24y 6z 42 31y 62 Add. x y z 1 7y 6z 20 31y 62
SECTION 10.1 Systems of Linear Equations: Sustitution and Elimination 735 We now solve equation for y y dividing oth sides of the equation y -31. x + y - z = c -7y + 6z = y = -1 20-2 Back-sustitute y = -2 in equation and solve for z. -7y + 6z = 20-71-22 + 6z = 20 6z = 6 z = 1 Sustitute y = -2 in. Sutract 14 from oth sides of the equation. Divide oth sides of the equation y 6. Finally, we ack-sustitute y = -2 and z = 1 in equation and solve for x. x + y - z = -1 x + 1-22 - 1 = -1 x - 3 = -1 x = 2 Sustitute y = -2 and z = 1 in. Simplify. Add 3 to oth sides. The solution of the original system is x = 2, y = -2, z = 1. You should verify this solution. Look ack over the solution given in Example 9. Note the pattern of removing one of the variales from two of the equations, followed y solving this system of two equations and two unknowns. Although which variales to remove is your choice, the methodology remains the same for all systems. NOW WORK PROBLEM 43. 6 Identify Inconsistent Systems of Equations Containing Three Variales EXAMPLE 10 An Inconsistent System of Linear Equations 2x + y - z = -2 c x + 2y - z = -9 x - 4y + z = 1 Our plan of attack is the same as in Example 9. However, in this system, it seems easiest to eliminate the variale z first. Do you see why? Multiply each side of equation y -1 and add the result to equation.add equations and. 2x y z 2 x 2y z 9 x y 7 x 2y z 9 x 4y + z 1 2x 2y 8 Multiply y 1. Add. Add. 2x y z 2 x y 7 2x 2y 8
736 CHAPTER 10 Systems of Equations and Inequalities x y 7 2x 2y 8 We now concentrate on equations and, treating them as a system of two equations containing two variales. Multiply each side of equation y 2 and add the result to equation. Multiply y 2. 2x 2y 14 2x 2y 8 0 22 Add. Equation has no solution and the system is inconsistent. 2x y z 2 x y 7 0 22 EXAMPLE 11 7 Express the of a System of Dependent Equations Containing Three Variales Solving a System of Dependent Equations x - 2y - z = 8 c 2x - 3y + z = 23 4x - 5y + 5z = 53 x 2y z 8 2x 3y z 23 x 2y z 8 4x 5y 5z 53 Our plan is to eliminate x from equations and. Multiply each side of equation y -2 and add the result to equation.also, multiply each side of equation y -4 and add the result to equation. Multiply y 2. 2x 4y 2z 16 2x 3y z 23 y 3z 7 Multiply y 4. 4x 8y 4z 32 4x 5y 5z 53 3y 9z 21 Add. Treat equations and as a system of two equations containing two variales, and eliminate the variale y y multiplying oth sides of equation y -3 and adding the result to equation. y 3z 7 Multiply y 3. 3y 9z 21 x 2y z 8 3y 9z 21 3y 9z 21 Add. y 3z 7 0 0 0 0 The original system is equivalent to a system containing two equations, so the equations are dependent and the system has infinitely many solutions. If we solve equation for y, we can express y in terms of z as y = -3z + 7. Sustitute this expression into equation to determine x in terms of z. x - 21-3z + 72 - z = 8 x + 6z - 14 - z = 8 We will write the solution to the system as where z can e any real numer. x - 2y - z = 8 x + 5z = 22 Add. x = -5z + 22 x = -5z + 22 e y = -3z + 7 x 2y z 8 y 3z 7 3y 9z 21 Sustitute y = -3z + 7 in. Remove parentheses. Comine like terms. Solve for x.
SECTION 10.1 Systems of Linear Equations: Sustitution and Elimination 737 This way of writing the solution makes it easier to find specific solutions of the system. To find specific solutions, choose any value of z and use the equations x = -5z + 22 and y = -3z + 7 to determine x and y. For example, if z = 0, then x = 22 and y = 7, and if z = 1, then x = 17 and y = 4. NOW WORK PROBLEM 45. Two points in the Cartesian plane determine a unique line. Given three noncollinear points, we can find the (unique) quadratic function whose graph contains these three points. EXAMPLE 12 Curve Fitting Find real numers a,, and c so that the graph of the quadratic function y = ax 2 + x + c contains the points 1-1, -42, 11, 62, and 13, 02. We require that the three points satisfy the equation y = ax 2 + x + c. For the point 1-1, -42 we have: -4 = a1-12 2 + 1-12 + c -4 = a - + c For the point 11, 62 we have: 6 = a112 2 + 112 + c 6 = a + + c For the point 13, 02 we have: 0 = a132 2 + 132 + c 0 = 9a + 3 + c We wish to determine a,, and c so that each equation is satisfied. That is, we want to solve the following system of three equations containing three variales: Figure 6 10 4 7 50 a - + c = -4 c a + + c = 6 9a + 3 + c = 0 Solving this system of equations, we otain a = -2, = 5, and c = 3. So the quadratic function whose graph contains the points 1-1, -42, 11, 62, and 13, 02 is y = -2x 2 + 5x + 3 y = ax 2 + x + c, a = -2, = 5, c = 3 Figure 6 shows the graph of the function along with the three points.