MA 15800, Summer 2016 Lesson 25 Notes Solving a System of Equations by substitution (or elimination) Matrices. 2 A System of Equations

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1 MA 800, Summer 06 Lesson Notes Solving a Sstem of Equations b substitution (or elimination) Matrices Consider the graphs of the two equations below. A Sstem of Equations From our mathematics eperience, ou probabl know that the first equation represents a parabola and the second equation represents a line. If the graphs of these two equations intersect, the ordered pair(s) is(are) called the solution(s) of the sstem of equations. Below are the graphs of both equations with the two solutions shown. The graph of the first equation, the parabola, is in black and the graph of the second equation, the line, is in red (, ) (, ) The ordered pairs above not onl represent the points where the graphs of the two equations intersect, the are also the onl ordered pairs that satisf both equations. In other words, the are the onl ordered pairs that make both equations true when the variables and are replaced. Using the point (, ): Using the point (, ) ( ) () ( ) A solution to a sstem of two equations makes both equations true. Hint: Before solving a sstem of equations alwas think about how man solutions there ma be. For eample: A line and a parabola could intersect at two points (as above) or onl one point or at no points (empt set).

2 Substitution Method:. Solve one of the equations for one variable in terms of the other variable.. Substitute the epression found in step one into the other equation, obtaining an equation of one variable.. Solve for this variable.. Back-substitute into the simplest equation to find the remaining variable(s).. Check each ordered pair in the given sstem E : Solve the sstem below. Write the solution(s) as ordered pair(s). Verif the solution b graphing each equation. I will solve the first equation for and substitute that epression for in the second equation. ( ) 7 ( ) Solution: (, ) Computerized graph: Intersection of the two lines is the solution First equation s graph is in black; the second equation s graph is in red. Intersection of the two lines is (, -) **Note: When using the substitution method and both variables drop out when solving and a false statement, such as =, results, there is no solution. In other words, the graphs of these two equations do not intersect and there are no ordered pair that will satisf both equations.**

3 Solve the following sstems of equations using the substitution method. E : Note: When solving this sstem of equations, the quadratic equation will be used. 0 Substitute the for the in the bottom equation. 0 I will use the quadratic formula to solve. ()() 8 () There are no real solutions to this equation. Therefore, this sstem of equations has NO SOLUTION. Verif b graphing that the graphs of these two equations do not intersect. If there is no solution, the graphs of the two equations will not intersect. Below are the graphs of both equations. The top equation is in black; it is a parabola opening right. The bottom equation, a line, is in red

4 With a sstem of two linear equations, the elimination method ma be used.. If necessar, multipl one of the equations b a non-zero constant so that the coefficients of one of the variables are opposites.. Add the two equations together and solve for one variable.. Back-substitute to find the remaining variable.. Check our ordered pair in both equations. E : Solve, using the elimination method. 6 The coefficients of the s are alread opposites. I will add the two equations and solve for. 6 7 Back-substitute in the first equation and solve for. Solution: (, ) The graph below verifies that the solution (intersection of the two lines) is (, ). The first equation is in black and the second in red

5 We have discussed that there is no solution if both variables drop out and a false statement results. When using the elimination method, if both variables drop out and a true statement results ( =, 0 = 0, etc.), then there are an infinite number of solutions. This result indicates that the two equations represent the same line. The solution is written in the form {(, ) equation of line}. Do not write all real numbers. Solutions are ordered pairs, not numbers. Graphicall, this means that the equations represent the same line. E. Solve using the substitution method. Substitue for in the other equation. () As seen above, both variables were eliminated and a false statement resulted. This indicates that there is NO SOLUTION and the graph would be parallel lines. The graph of the lines is seen below. The black line is the first equation; red is the second E : Solve using the elimination method. 6 ( 6) Add the two equations together: 0 0 As seen above, both variables were eliminated but a TRUE statement resulted. This indicates that there are an INFINITE number of solutions because the two equations are the same line. {(, ) } or The solution would be written {(, ) 6} or {(, )} 8

6 Characteristics of a Sstem of Linear Equations in Variable Graphs Number of Solutions Classification Nonparallel Lines One Consistent and Independent Sstem Same Lines Infinite Consistent and Dependent Sstem Parallel Lines No Solution Inconsistent Sstem E : Classif the following sstems of linear equations as consistent and independent, consistent and dependent, or inconsistent. 7 c d a) b) 9 c d (7 ) 8 7( 9) 6 a) 9 There is a solution. Therefore the sstem is consistent and independent. Graph of sstem: First equation is in black, second in red b) First I will 'clear' the denominators in the equations. 6( c d ) c d 0 (c d 0) ( c d ) c d (c d ) c6d 60 9c 6d 9 c c 6

7 There is a solution (variables were not eliminated). Therefore the sstem is consistent and independent and the graph would be two nonparallel lines. Below is a graph of both lines. d c ( ) 8 c) 07 Variables were eliminated and a false statement resulted. This sstem has no solution. The graph would be two parallel lines. Solve each sstem of equations. If it is a linear sstem, describe its classification. 6 E 6: I will solve the second equation for and substitute that epression for the in the first equation. ( ) (6) 0 0 or , 6 (0) Solutions: (, 0), The first equation is a circle and the second is a line. The graphs are shown on the net page. 7

8 (, 6 ) (,0) E 7: I will substitute the epression for from the bottom equation for the in the top equation. 8 0 (8) 0 0 or The top equation is a parabola. The bottom equation is a cubic curve. Both are graphed below. 8 (0 ) 0 Solutions: (0,0),, ( 8, 8 ) 0.00 (0,0)

9 We have solved applied problems where onl one equation of one variable was used. Sometimes writing two equations in two variables can be used to solve applied problems. E 8: The price of admission to a high school pla was $ for students and $.0 for nonstudents. If 0 tickets were sold for a total of $.0, how man of each kind of tickets were purchased? Let = the number of student tickets sold Let = the number of nonstudent tickets sold One equation: + = 0 Another equation: +. =. I will use the elimination method, multipling the first equation b then adding so the terms are eliminated. ( 0) student tickets and 7 nonstudent tickets E 9: A rancher is preparing an oat-cornmeal miture for livestock. Each ounce of oats provided grams of protein and 8 grams of carbohdrates. An ounce of cornmeal provides grams of protein and grams of carbohdrates. How man ounces of each can be used to meet the nutritional goals of 00 grams of protein and 0 grams of carbohdrates per feeding? Let a = number of ounces of oats Let b = number of ounces of cornmeal One equation: a + b = 00 Second equation: 8a + b = 0 I will use the elimination method, multipling the first equation b 8 and leaving the second equation as is. 8(a b 00) a b 600 8a b 0 8a b 0 a 80 a 0 ab00 (0) b b 00 b 0 b 0 0 ounces of oats and 0 ounces of cornmeal 9

10 E 0: Solve the following sstem. ( ) ( ) 0 ( ) ( ) 0 ( ) ( ) ( ) 0 0 or 0 0 Substitute in top equation. 0 Solutions: (0,), (, ) Both solutions check in both equations. The top equation is a circle with the center at (, ) and a radius length 0. The bottom equation is a line. Both are graphed below and the solutions ma be seen. (0, ) (, )