SOLVING SYSTEMS OF EQUATIONS

Transcription

1 SOLVING SYSTEMS OF EQUATIONS In this course, one focus is on what a solution means, both algebraicall and graphicall. B understanding the nature of solutions, students are able to solve equations in new and different was. This understanding also provides opportunities to solve some challenging application problems. In Chapter this knowledge is etended to solve equations and sstems of equations with three variables. Eample The graph of = ( 5) 2 4 is shown at right. Use the graph to solve each of the following equations. a. ( 5) 2 4 = 2 b. ( 5) 2 4 = 3 c. ( 5) 2 = 4 Solutions: a. Add the graph of = 2 which is a horizontal line to the graph of = ( 5) 2 4. These two graphs intersect at two points, (, 2) and (9, 2). The -coordinates of these points are the solutions to the original equation. Notice that there is no in the equation in part (a). Therefore the solutions to the equation are = and = 9. b. Add the graph of = 3 to see that the graphs intersect at (4, 3) and (6, 3). Therefore the solutions to are = 4 and = 6. c. The equation might look as if it cannot be solved with the graph, but it can. B recognizing the equation is equivalent to ( 5) 2 4 = 0 (subtract 4 from both sides), then the graph can be used to find where the parabola crosses the line = 0 (the -ais). The graph tells us the solutions are = 7 and = 3. Parent Guide and Etra Practice 205 CPM Educational Program. All rights reserved. 27

2 Eample 2 Solve the equation + 2 = 2 + using at least two different methods. Eplain our methods and the implications of the solution(s). Solution: One method is to use algebra to solve this equation. This involves squaring both sides and solving a quadratic equation as shown at right. A problem arises, however, if the solutions are not checked. When each -value is substituted back into the original equation, onl one -value checks: =. This is the onl solution = 2 + ( + 2 ) 2 = (2 + ) = = 0 (4 )( + ) = 0 = 4, = 2 ( 4 ) = = = 2( ) + = 2 + To see wh the other solution does not work use a graph to solve the equation. The graphs of = + 2 and = 2 + are shown at right. Notice that the graphs onl intersect at one point, namel =. There is onl solution to the equation; the other 4 solution is called an etraneous solution. Remember that a solution makes the equation true. In the original equation, this means that both sides of the equation will be equal for certain values of. Using the graphs, the solution is the -value that has the same -value for both functions, or the -coordinate(s) of the point(s) at which the graphs intersect CPM Educational Program. All rights reserved. Core Connections Integrated III

3 Eample 3 Algebraicall solve each sstem of equations below. For each sstem, eplain what the solution (or lack thereof) tells about the graph of the sstem. a. = b. = 2( 2) = = c. = = 25 Solutions: a. The two equations are written in = form, which makes substitution the most efficient method for solving. Set the epressions on the right side of each equation equal to each other and solve for. Then substitute this value for back into either one of the original equations to determine the value of. Finall, check that the solution satisfies both equations. = = = = Check ( ) = 5 ( ) 2 +5 = = 25 = 5 Solution to check: (5, ) The solution is the point (5, ), which means that the graphs of these two equations intersect at one point, the point (5, ). b. The two equations are written in = form, which means that substitution can again be used. This is shown at right. Now substitute each -value into either equation to calculate the corresponding -value. = 6, = = 2(6) + 5 = = 3 Solution: (6, 3) =, = = 2( ) + 5 = = 7 Solution: (, 7) 2( 2) = ( 2) 2 = ( ) = = = 0 2( 2 5 6) = 0 2( 6)( + ) = 0 2 0, = 6, = Solution continues on net page Parent Guide and Etra Practice 205 CPM Educational Program. All rights reserved. 29

4 Solution continued from previous page. Lastl, check each point in both equations to make sure there are not an etraneous solutions. (6, 3): = 2( 2) = 2(6 2) = 2(6) + 35 (6, 3): = = 2(6) + 5 (, 7): = 2( 2) = 2( 2) = 2(9) + 35 (, 7): = = 2( ) + 5 In solving these two equations with two unknowns, two solutions were found, both of which checked in the original equations. This means that the graphs of the equations, a parabola and a line, intersect in two distinct points. c. This sstem requires substitution to solve. One option is to replace in the second equation with the right hand side of the first equation, but that would require solving an equation of degree four (an eponent of 4). Instead, rewrite the first equation without fractions in order to simplif. This is done b = multipling both sides of the equation b 6, as shown at right. 6 = 2 34 Now, instead, replace the 2 in the second equation with Then solve. Net, substitute this value back into either equation to find the corresponding -value = 25 (6 + 34) + 2 = = = 0 ( + 3)( + 3) = 0 = = 2 = 3: = 2 6( 3) + 34 = = 2 6 = 2 = ±4 This gives us two possible solutions: (4, 3) and ( 4, 3). Be sure to check these points for etraneous solutions! (4, 3) : = , 3 = 6 (4) = = 8 6 (4, 3) : = 25, (4) 2 + ( 3) 2 = = 25 ( 4, 3) : = , 3 = 6 ( 4) = = 8 6 ( 4, 3) : = 25, ( 4) 2 + ( 3) 2 = = 25 Since there are two points that make this sstem true, the graphs of this parabola and this circle intersect in onl two points, (4, 3) and ( 4, 3) CPM Educational Program. All rights reserved. Core Connections Integrated III

5 Eample 4 Jo has small containers of lemonade and lime soda. She once mied one container of lemonade with three containers of lime soda to make 7 ounces of a tast drink. Another time, she combined five containers of lemonade with si containers of lime soda to produce 58 ounces of another splendid beverage. Given this information, how man ounces are in each small container of lemonade and lime soda? Solution: Solve this problem b using a sstem of equations. To start, let = the number of ounces of lemonade in each small container, and let = the number of ounces of lime soda in each of its small containers. Write an equation that describes each miture Jo created. The first miture used one container ( ounces) of lemonade and three containers (3 ounces) of lime soda for a total of 7 ounces. This can be represented as + 3 = 7. The second miture used five containers (5 ounces) of lemonade and si containers (6 ounces) of lime soda for a total of 58 ounces. This can be represented b the equation = 58. Solve this sstem to determine the values of and. + 3 =7 ( 5) 5 5 = = = 58 (Note: Check these values!) 9 = 27 = 3 If = 3, then: + 3(3) =7 + 9 =7 = 8 Therefore each container of lemonade has 8 ounces, and each container of lime soda has 3 ounces. Parent Guide and Etra Practice 205 CPM Educational Program. All rights reserved. 3

6 Problems Solve each of the following sstems of equations. Then eplain what the solution(s) tells ou about the graphs of the equations. Be sure to check our work.. + = 3 = = = = = = = = 24 = = = 24 The graph of = 2 ( 4)2 + 3 is shown at right. Use the graph to solve each of the following equations. Eplain how ou get our answers ( 4)2 + 3 = ( 4)2 + 3 = ( 4)2 + 3 = 0. 2 ( 4)2 = 8 Solve each equation below.. 3( 4) = = ( ) = = 0 Solve each of the following sstems of equations algebraicall. What does the solution tell ou about the graph of the sstem? 5. = = = ( + ) = = 3( 4) 2 2 = = 0 = ( 4) Adult tickets for the Mr. Moose s Fantas Show on Ice are $6.50 while a child s ticket is onl $2.50. At Tuesda night s performance, 435 people were in attendance. The bo office brought in $ for that evening. How man of each tpe of ticket were sold? 20. The net math test will contain 50 questions. Some will be worth three points while the rest will be worth si points. If the test is worth 95 points, how man three-point questions are there, and how man si-point questions are there? CPM Educational Program. All rights reserved. Core Connections Integrated III

7 2. Dudle s water balloons follow the path described b the equation = 8 25 ( 0) Suppose Dudle s nemesis, in a mad dash to save his base from total water balloon bombardment, ran to the wall and set up his launcher at its base. Dudle s nemesis launches his balloons to follow the path = ( ) in an effort to knock Dudle s water bombs out of the air. Is Dudle s nemesis successful? Eplain. 5. Answers. (4, 7) 2. ( 2, 5) 3. no solution 4. ( 2, ) 5. all real numbers 6. (2, 3) 7. = 4 The horizontal line = 3 crosses the parabola at one point, at the verte. 9. no real solution The horizontal line = does not cross the parabola. (Solving algebraicall ields = 4 ± 2i.) 8. = 2 or = 6 The horizontal line = 5 crosses the parabola at two points. 0. = 0 or = 8 Add three to both sides to rewrite the equation as 2 ( 4)2 + 3 =. The horizontal line = crosses the parabola at two points.. = 7 or = 2. no solution 3. = 2 4. no solution 5. all real numbers When graphed, these equations give the same line. 7. no solution This parabola and this line do not intersect adult tickets were sold, while 290 child tickets were sold. 6. (0, 4) The parabola and the line intersect at one point. 8. (2, 2) and (5, 5) The line and the parabola intersect twice. 20. There are 35 three-point questions and 5 si-point questions on the test. 2. B graphing we see that the nemesis balloon when launched at the base of the wall (the -ais), hits the path of the Dudle s water balloon. Therefore, if timed correctl, the nemesis is successful. Parent Guide and Etra Practice 205 CPM Educational Program. All rights reserved. 33