SOLVING SYSTEMS OF EQUATIONS 4.. 4..4 Students have been solving equations even before Algebra. Now the focus 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. Their understanding also provides opportunities to solve some challenging applications. In this section the will etend their knowledge about solving one and two variable equations to solve sstems with three variables. Eample The graph of = (! 5) 2! 4 is shown below right. Solve each of the following equations. Eplain how the graph can be used to solve the equations. a. (! 5) 2! 4 = 2 b. (! 5) 2! 4 =!3 c. (! 5) 2 = 4 Students can determine the correct answers through a variet of different was. For part (a), most students would do the following: (! 5) 2! 4 = 2 (! 5) 2 = 6! 5 = ±4 = 5 ± 4 = 9, This is correct and a standard procedure. However, with the graph of the parabola provided, the student can find the solution b inspecting the graph. Since we alread have the graph of = (! 5) 2! 4, we can add the graph of = 2 which is a horizontal line. These two graphs cross at two points, and the -coordinates of these points are the solutions. The intersection points are (, 2) and (9, 2). Therefore the solutions to the equation are = and = 9. We can use this method for part (b) as well. Draw the graph of = 3 to find that the graphs intersect at (4, 3) and (6, 3). Therefore the solutions to part (b) are = 4 and = 6. The equation in part (c) might look as if we cannot solve it with the graph, but we can. Granted, this is eas to solve algebraicall: (! 5) 2 = 4! 5 = ±2 = 7, 3 But, we want students to be looking for alternative approaches to solving problems. B looking for and eploring alternative solutions, students are epanding their repertoire for solving problems. This ear the will encounter equations that can onl be solved through alternative methods. B recognizing the equation in part (c) as equivalent to (! 5) 2! 4 = 0 (subtract four from both sides), we can use the graph to find where the parabola crosses the line = 0 (the -ais). The graph tells us the solutions are = 7 and = 3. 40 Core Connections Algebra 2
Chapter 4 Eample 2 Solve the equation + 2 = 2 + using at least two different approaches. Eplain our methods and the implications of the solution(s). Most students will begin b using algebra to solve this equation. This includes squaring both sides and solving a quadratic equation as shown at right. A problem arises if the students do not check their work. If the students substitute each -value back into the original equation, onl one -value checks: =. This is the onl solution. We 4 can see wh the other solution does not work if we 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 =. This is the onl point where the are 4 equal. The solution to the equation is this one value; the other value is called an etraneous solution. Remember that a solution is a value that makes the equation true. In our original equation, this would mean that both sides of the equation would be equal for certain values of. Using the graphs, the solution is the -value that has the same -value for both graphs, or the point(s) at which the graphs intersect. + 2 = 2 + ( + 2 ) 2 = (2 + ) 2 + 2 = 4 2 + 4 + 4 2 + 3! = 0 (4! )( + ) = 0 = 4,! Eample 3 Solve each sstem of equations below without graphing. For each one, eplain what the solution (or lack thereof) tells ou about the graph of the sstem. a. =! 2 5 + 3 b. =!2(! 2)2 + 35 = 3 5! 2 =!2 + 5 c. = 6 2! 34 6 2 + 2 = 25 The two equations in part (a) are written in = form, which makes the Substitution method the most efficient method for solving. Since both epressions in are equal to, we set the epressions equal to each other, and solve for. =! 2 5 + 3 = 3 5! 2 =! 2 5 + 3 = 3 5! 2 ( ) = 5 ( 5 3! 2 )!2 +5 = 3!0 5! 2 5 + 3!5 =!25 = 5 Parent Guide and Etra Practice 4
We substitute this value for back into either one of the original equations to determine the value of. Finall, we must check that the solution satisfies both equations. Check Therefore the solution is the point (5, ), which means that the graphs of these two lines intersect in one point, the point (5, ). The equations in part (b) are written in the same form so we will solve this sstem the same wa we did in part (a). We substitute each -value into either equation to find the corresponding -value. Here we will use the simpler equation. = 6, =!2 + 5 =!2(6) + 5 =!2 + 5 = 3 Solution to check: (5, ) Solution: (6, 3) =!, =!2 + 5 =!2(!) + 5 = 2 + 5 = 7 Solution: (!, 7) Lastl, we need to check each point in both equations to make sure we do not have an etraneous solutions. =!2(! 2) 2 + 35 =!2 +5!2(! 2) 2 =!2! 20!2( 2! 4 + 4) =!2! 20!2 2 + 8! 8 =!2! 20!2 2 +0 +2 = 0 2! 5! 6 = 0 (! 6)( +) = 0 = 6, =! (6, 3) : =!2(! 2) 2 + 35 (6, 3) : =!2 +5 3 =!2(6! 2) 2 + 35 3 =!2(6) +5 Check 3 =!2(6) + 35 Check (!, 7) : =!2(! 2) 2 + 35 7 =!2(!! 2) 2 + 35 7 =!2(9) + 35 Check (!, 7) : =!2 +5 7 =!2(!) +5 Check In solving these two equations with two unknowns, we found two solutions, both of which check in the original equations. This means that the graphs of the equations, a parabola and a line, intersect in eactl two distinct points. 42 Core Connections Algebra 2
Chapter 4 Part (c) requires substitution to solve. We can replace in the second equation with what the first equation tells us it equals, but that will require us to solve an equation of degree four (an eponent of 4). Instead, we will first rewrite the first equation without fractions in an effort to simplif. We do this b multipling both sides of the equation b si. = 6 2! 34 6 6 = 2! 34 6 + 34 = 2 Now we can replace the 2 in the second equation with 6 + 34. Net we substitute this value back into either equation to find the corresponding -value. 2 + 2 = 25 (6 + 34) + 2 = 25 2 + 6 + 34 = 25 2 + 6 + 9 = 0 ( + 3)( + 3) = 0 =!3 =!3 : 6 + 34 = 2 6(!3) + 34 = 2!8 + 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) : = 6 2! 34 6,! 3 = 6 (4)2! 34 6 = 6 6! 34 6 =!8 6, check. (4,!3) : 2 + 2 = 25, (4) 2 + (!3) 2 = 6 + 9 = 25, check. (!4,!3) : = 6 2! 34 6,! 3 = 6 (!4)2! 34 6 = 6 6! 34 6 =!8 6, check. (!4,!3) : 2 + 2 = 25, (!4) 2 + (!3) 2 = 6 + 9 = 25, check. 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). Parent Guide and Etra Practice 43
Eample 4 Jo has small containers of lemonade and lime soda. She once mied one lemonade container 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? We can solve this problem b using a sstem of equations. To start, we let equal the number of ounces of lemonade in each small container, and equal the number of ounces of lime soda in each of its small containers. We can write an equation that describes each miture Jo created. The first miture used one ounce of liquid and three ounces of liquid to produce 7 ounces. This can be represented as + 3 = 7. The second miture used five ounces of liquid and si ounces of liquid to produce 58 ounces. This can be represented b the equation 5 + 6 = 58. We can solve this sstem to find the values for and. + 3 = 7! "5 5 + 5 = 85 5 + 6 = 58! 5 + 6 = 58 + 3(3) = 7 + 9 = 7 = 8 9 = 27 = 3 (Note: ou should check these values!) Therefore each container of Jo s lemonade has 8 ounces, and each container of her lime soda has onl 3 ounces. 44 Core Connections Algebra 2
Chapter 4 Problems Solve each of the following sstems for and. Then eplain what the answer tells ou about the graphs of the equations. Be sure to check our work.. + = 3! = 5 2. 2! 3 =!9!5 + 2 = 20 3. 5 + 0 = 2 6 + 4 = 4. 8 + 2 = 8!6 + = 4 5. 2! 6 = 24 = 3 4! 3 2 6.! 7 =!5 2 3! 4 = 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 answer. 7. 2 (! 4)2 + 3 = 3 8. 2 (! 4)2 + 3 = 5 9. 2 (! 4)2 + 3 = 0. 2 (! 4)2 = 8 Solve each equation below. Think about rewriting, looking inside, or undoing to simplif the process.. 3(! 4) 2 + 6 = 33 2. 4 + 5 = 9!4 20 3. 3 + ( 0!3 2 ) = 5 4.!3 2! + 4 = 0 Solve each of the following sstems of equations algebraicall. What does the solution tell ou about the graph of the sstem? 5. =! 2 3 + 7 4 + 6 = 42 6. = ( + ) 2 + 3 = 2 + 4 7. =!3(! 4) 2! 2 =! 4 7 + 4 8. + = 0 = (! 4) 2! 6 9. 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 $667.50 for that evening. How man of each tpe of ticket were sold? Parent Guide and Etra Practice 45
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? 2. Reread Eample 3 from Chapter 4 about Dudle s water balloon fight. If ou did this problem, ou found that Dudle s water balloons followed the path described b the equation =! 8 25 (! 0)2 + 72. Suppose Dudle s nemesis, in a mad dash to save his 5 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 =! (! 89 25 ) in an effort to knock Dudle s water bombs out of the air. Is Dudle s nemesis successful? Eplain. 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 in onl one point, at the verte. 9. No solution. The horizontal line = does not cross the parabola. 8. = 2, = 6 0. = 0, = 8. Add three to both sides to rewrite the equation as 2 (! 4)2 + 3 =. The horizontal line = crosses at these two points.. = 7, = 2. no solution 3. = 2 4. No solution. (A square root must have a positive result.) 5. All real numbers. When graphed, these equations give the same line. 7. No solution. This parabola and this line do not intersect. 9. 45 adult tickets were sold, while 290 child tickets were sold. 6. (0, 4). The parabola and the line intersect onl once. 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. 46 Core Connections Algebra 2