# Systems of linear equations (simultaneous equations)

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1 Before starting this topic ou should review how to graph equations of lines. The link below will take ou to the appropriate location on the Academic Skills site. The topics ou need to review are listed below. Topic 1: Eamples of linear relationships Topic : Graphing lines Topic 3: Finding equations of lines Sstems of linear equations (simultaneous equations) The eample below will serve as an introduction to this module. Eample; If the ABC Tai compan charges a flag fall of \$ plus \$1 per km, this can be modelled b the linear equation: C = 1k+ where C is the cost to travel k kilometres. Another tai compan has a charging structure which gives the equation C = 1.k. In this eample there are two equations containing two variables (cost, kilometres). Graphing these together gives 0 C C=1.k C=1k k This is called a sstem of linear equations. The point where the lines intersect (cross) is the solution of the sstem. From the graph it is possible to see that the fare will be the same (\$1) for a km journe. Centre for Teaching and Learning Academic Practice Academic Skills Page 1

2 There are three methods of solving a sstem of linear equations presented in this module; i. Graphicall ii. iii. Algebraicall Determinants There is another method using Matrices (See Matrices module). This module will be available in earl 014. This module uses onl sstems of two equations with two unknowns. A more comple sstem of linear equations would involve 3 equations with 3 variables. Each of the methods mentioned above can possibl be used to solve more comple sstems, however, the use of determinants and matrices becomes the preferred method. Centre for Teaching and Learning Academic Practice Academic Skills Page

3 Graphicall In this section, the two lines are drawn on the same set of aes. Unless the lines are parallel, the lines will intersect. This point is called the point of intersection. It is a point that satisfies both equations. One particular problem with this method is that the point of intersection ma onl be located approimatel. Eample: Find the point of intersection of the two equations: = + 9 and =. In the table below, each line is shown separatel with the appropriate working. The final graph combines the two and the point of intersection determined. = + 9 = 3 intercept (when = 0) 0= + 9 = 9 intercept is (9,0) intercept (when = 0) = 0+ 9 = 9 intercept is (0,9) intercept (when = 0) 0= 3 = 3 intercept is (3,0) intercept (when = 0) = 0 3 = 3 intercept is (0,-3) Centre for Teaching and Learning Academic Practice Academic Skills Page 3

4 Combining the two gives: (6,3) The point of intersection is (6,3). The solution to the sstem of linear equations is (6,3). The point must also satisf both equations. To check this, the solution is substituted into both equations as shown below. In the equation = + 9, 6= 3+ 9; and in = 3, 3= 6 3, both equations are satisfied. Eample: Solve the sstem of equations; + = and 3 = 6. Note these equations are presented in general form. In the previous question the equations were presented in point-slope form. + = 3 = 6 intercept (when = 0) intercept (when = 0) intercept (when = 0) intercept (when = 0) + 0 = = intercept is (,0) 0 + = = intercept is (0,) 3 = 6 = intercept is (,0) 3 0 = 6 = 6 intercept is (0,-6) Centre for Teaching and Learning Academic Practice Academic Skills Page 4

5 Combining the two gives: 3 = = The point of intersection is approimatel (.3, 1.1). Remember, this solution is approimate. A more accurate wa of determining the point of intersection is b using algebra. Checking the solution in the first equation: + =, = 1., this close the epected value of. In the second equation: 3 = 6, = 8., this is close to the epected value of 6. It is worth remembering that the accurac of the point of intersection is limited b the abilit to determine the point on the graph. Centre for Teaching and Learning Academic Practice Academic Skills Page

6 There are two sstems that do not result in a single point of intersection. The first is called the Inconsistent sstem. The inconsistent sstem results in no solution(s). For eample: Consider the equations: = = 0 The graphs of these equations are: = = 0 - From the graph it is clear that the lines will never intersect. This can be shown b writing both equations in the point slope form. = 3+ 1 is alread in the point slope form. The slope of the line is = 0 rearranged into the point slope form becomes = 3 +. The slope of this line is also 3. As both slopes are 3, the lines are parallel; there will never be a point of intersection. The second sstem is the Dependent sstem. Like the eample above, it is a result of parallel lines. This time the lines are identical and have ever point in common. Centre for Teaching and Learning Academic Practice Academic Skills Page 6

7 The eample below shows this. Consider the equations: = = 0 The graph of these equations is: Writing the second equation in the point slope form reveals what is happening here. Both equations are = + 8. This indicates that the coordinates of both lines are the same. This eplains the absence of a single solution. Video Sstems of linear equations (Graph) Centre for Teaching and Learning Academic Practice Academic Skills Page 7

8 Algebraicall In this section, the solution to the sstem of linear equations is found algebraicall. The accurac of the solution is improved. Within this section, there are two basic strategies that can be used. The first is substituting one equation into the other. The second is using either addition or subtraction to eliminate one of the variables. An earlier eample is solved below. Substitution Eample: The two equations being considered are: = + 9 and =. With both equations are presented in the point slope form, the second equation can be substituted into the first. Substituting = into = + 9 becomes = + 9 = = 9 = 3 Now it is known that the value of is 3, the value of must be determined. This process is performed b substituting =3 into (sa) the first equation. Now the value of and is known, checking should be performed in the second (other) equation. Substituting = 3 into the first equation gives: = + 9 = 3+ 9 = 6 The solution is 3, 6 Checking in the second equation: = = 3 = 6 Solving this question was ver straight-forward. The graphical method gives a good picture of what is happening; however, the algebraic method is often quicker. Eample: Solve the sstem of equations; + = and 3 = 6. This time both equations are presented in general form. To substitute one equation into the other, one equation will require rearranging. The second equation contains a single term, so it is the easiest to rearrange. Centre for Teaching and Learning Academic Practice Academic Skills Page 8

9 3 = 6 becomes = 3 6 Substituting this into the first equation gives + = ( ) = = 17 = =. 3 to d.p. 17 Taking the value of and substituting into the first equation to obtain should be performed using the eact value of. + = 40 + = = = = Checking in the second equation using eact values: = = = 8 = = 6 8 The check confirms the solution obtained. The solution is 1.06) , 17 8 which approimates to (.3, Centre for Teaching and Learning Academic Practice Academic Skills Page 9

10 Addition or subtraction (the elimination method) Eample: The two equations being considered are: = + 9 and =. The first step in this method is to write both equations in the same form, one above the other lining up like variables. In this eample this is alread done. = + 9 = The aim of this step is to eliminate one of the variables. If the equations were subtracted, the variable would be eliminated. = + 9 = 0= 3+ 9 = 3 The net steps are the same as previousl performed. Substituting = 3 into the first equation gives: = + 9 = 3+ 9 = 6 (, ) The solution is 3 6 Checking in the second equation: = 6= 3 Eample: The two equations being considered are: + = 7 and = 8. The first step in this method is to write both equations in the same form. + = 7 + = 7 = 8 + = 8 The term onl differs b sign, so if the two equations were added ( + - = 0). + = = 8 3 = 1 = Centre for Teaching and Learning Academic Practice Academic Skills Page

11 Substituting = into the first equation gives: Checking in the second equation: + = 7 + = 7 = 3 = 3 = 8 Eample: Find the point of intersection of the lines: = 7 and = 1 Write the equations in the same form. = 7 + = 7 = 1 = 1 In this eample, addition or subtraction cannot take place until the or coefficients are the same (ecept for sign). To achieve this, the first equation is multiplied b. After this the equations can be subtracted to eliminate the variable. + = =-14 - = = 1 Substituting = -3 into the first equation gives: = 7 = 3 7 = 6 7 = 1 =-1 =-3 Checking in the second equation: = 1 3 = 1 Centre for Teaching and Learning Academic Practice Academic Skills Page 11

12 Eample: Find the point of intersection of the lines: 3+ = 9 and 4 3= 17 The equations are in the same form. This question requires the multiplication of both equations in order to eliminate a variable. To eliminate, the first equation is multiplied b 4 and the second b 3. To eliminate, the first equation is multiplied b 3 and the second b. Eliminating b subtracting: 3+ = = = = 1 Substituting = -3 into the first equation gives: Checking in the second equation: 3+ = = 9 3 = 6 = 4 3 = = 8+ 9 = 17 9= 87 = 3 Eample: Solve the sstem of equations; + = and 3 = 6. The equations are in the same form. If the second equation is multiplied b, then the variable can be eliminated b addition. + = + = 3 = = = = 17 Centre for Teaching and Learning Academic Practice Academic Skills Page 1

13 Substituting 40 = into the first equation gives: 17 Checking in the second equation: + = 40 + = = = = = = 8 = = 6 8 Video Sstems of linear equations (Algebraic) Video Sstems of linear equations (Application) Centre for Teaching and Learning Academic Practice Academic Skills Page 13

14 Using determinants In this section, a totall different method of solving Sstems of Linear Equations will be covered. The theor behind the method used will not be covered, but the process of how determinants can be used will be presented as a sequence of steps to follow. It is necessar to be able to solve a determinant of the second order. An eample of a second order determinant is: a b det ad bc c d = Evaluating a second order determinant is performed as below. det 4 1 = 4 7 ( 1) 3 = = Eample: Solve this sstem of linear equations using determinants: = 3+ and =. The first step is to write both equations in the form of a + b = c = = = + 1= 0 constants coefficients From this a (square) matri (A) will be formed using onl the coefficients of the variables and. A 3 1 = 1 This matri is modified to produce two new matrices. The matri A* is formed b replacing the coefficients with the constants. The matri A** is formed b replacing the coefficients with the constants. Altogether there are three matrices A = A* = A* * = Centre for Teaching and Learning Academic Practice Academic Skills Page 14

15 The solution of the sstem of linear equations is given b: 1 det A* = = = = = det A ( ) 1 3 det A** = = = = = 4 det A The solution (point of intersection) is (,4). This can be checked b substituting into the original equations as in previous methods. This method of solving linear equations is called Cramer s Rule. Once the method is practised, solving linear equations is fairl straightforward. Eample: Find the point of intersection of the lines: 3+ = 9 and 4 3= 17 The first step is to write both equations in the form of a + b = c 3+ = 9 4 3= 17 From this a square matri (A) will be formed using onl the coefficients of the variables and. A 3 = 4 3 Modif this matri to produce two new matrices. The matri A* is formed b replacing the coefficients with the constants. The matri A** is formed b replacing the coefficients with the constants. Altogether there are three matrices A = A* = A* * = Centre for Teaching and Learning Academic Practice Academic Skills Page 1

16 The solution of the sstem of linear equations is given b: 9 det A* = = = = = det A det A** = = = = = 3 det A The solution (point of intersection) is (,-3). This can be checked b substituting into the original equations as in previous methods. Eample: Solve the sstem of equations; + = and 3 = 6. Both equations are in the correct form. From this the matri A = = 3 = 6 From this the matrices A* and A** are formed. A = 3 1 A* = 6 1 A* * = 3 6 The solution is: det A* = = = = = det A det A** = = = = = det A ( 1) The solution (point of intersection on the graph) is, Centre for Teaching and Learning Academic Practice Academic Skills Page 16

17 Video Sstems of linear equations (Determinants) Centre for Teaching and Learning Academic Practice Academic Skills Page 17

18 Activit 1. Evaluate the determinants (a) (b) 0 (c) (d) Solve the sstem of linear equations b the method indicated. (a) = + 4 = 3 Graphical method (c) 4+ = 19 + = 8 Graphical method (e) 4+ = 19 + = 8 Algebraic method (g) = = 0 Algebraic method (i) 4+ 3 = 6 = Algebraic method (k) 4+ = 19 + = 8 Determinant method (m) = = 6. Determinant method (b) + = = Graphical method (d) 3+ = 7 3 = 1 Graphical method (f) 3+ = 7 3 = 1 Algebraic method (h) = = 6. Algebraic method (j) + 11 = = 0 Algebraic method (l) 4+ 3 = 6 = Determinant method (n) + 11 = = 0 Determinant method (o) (p) A customer purchases four chocolate bars and one ice cream and pas \$8.70; another customer purchases one chocolate bar and two ice creams and pas \$6., what is the unit price of chocolate bars and ice creams? The sum of two numbers is 4, where the difference of the two numbers is 9. Determine the two numbers. Centre for Teaching and Learning Academic Practice Academic Skills Page 18

19 (q) Two cables support a weight of 10N as shown in the diagram below. T 1 T 10N N The equations below appl to the direction indicated; Verticall T T = 10 Horizontall T 1 = + 0.T Calculate the tensions T 1 and T (r) A volume of 6% solution is mied with a different volume of 1% solution to obtain 00mL of a % solution. Calculate the volumes of each of the original solutions. Centre for Teaching and Learning Academic Practice Academic Skills Page 19

20 Answers 1. Evaluate the determinants (a) = = (b) = ( ) 0 = 0 0 (c) = ( 1. ) ( 1. ) 8= 1. 8 (d) 1 = ( 1)( + 1) = 1 = Solve the sstem of linear equations b the method indicated. (a) = + 4 = 3 Graphical method (b) + = = Graphical method (,6) (,0) (c) 4+ = 19 + = 8 Graphical method (d) 3+ = 7 3 = 1 Graphical method - - (-4.,-1) - - (-3.3,-1.) Centre for Teaching and Learning Academic Practice Academic Skills Page 0

21 (e) 4+ = 19 + = 8 Algebraic method 4+ = = 8 6 = -7 = = =-19 = -1 Solution is (-4.,-1) Check in equation - + (. ) = = 8 (g) = = 0 Algebraic method = = = = = 0 = = = 0 7 = 14 = Checking ( ) Solution is, = = 0 (f) 3+ = 7 3 = 1 Algebraic method 3+ = 7 3+ = 7 3 = = 3 7 = = 7 3+ = 7 3+ = = = 7 3 Solution is, 7 7 Checking 3 3 = = = 1 7 (h) = = 6. Algebraic method = = = = 1. ( 3) = 3. 4 = = = 18. Solution is, Checking = =. 6. = 6 = Centre for Teaching and Learning Academic Practice Academic Skills Page 1

22 (i) 4+ 3 = 6 = Algebraic method 4+ 3 = 6 = b substitution ( ) 4+ 3 = 6 = 1 6 = = = = = 6 Solution is, Checking =? 6 = = (k) 4+ = 19 + = 8 Determinant method A = 4 3 A* = 17 3 A* * = 4 17 det A* = = = = det A det A** = = = = 3 det A 9 9 The solution is (, 3) (m) = = 6. Determinant method A = A* A* * = = det A* = = = = det A det A** = = = = 3 det A The solution is, ( 3) (j) + 11 = = 0 Algebraic method + 11 = = = 0 + = 40 = = = ( 4 8) Solution is, Checking = = 10 = 4 = 0 (l) 4+ 3 = 6 = Determinant method 4+ 3 = = 6 = = A = 1 A* = A* * 1 = det A* = = = = det A 4 6 det A** = = = = det A 6 The solution is, (n) + 11 = = 0 Determinant method Centre for Teaching and Learning Academic Practice Academic Skills Page

23 + 11 = = = = A = A* A* * 0. 1 = 4 1 = 0. 4 det A* = = = = 4 det A. 0. det A** = = = = 8 det A The solution is, ( 4 8) (o) (p) A customer purchases four chocolate bars and one ice cream and pas \$8.70; another customer purchases one chocolate bar and two ice creams and pas \$6., what is the unit price of chocolate bars and ice creams? Let the price of the chocolate bar be C, let the price of the ice-cream be I. The two equations formed are: 4C+ I = 8. 7 and C + I = 6. Solving b algebra: 4C + I = 87. 4C + I = C + I = 6. 4C + 8I = 6. 7I = 17. I =. 4C +. = 87. 4C = 6. C = 1. Solution is Chocolate Bars cost \$1. and Ice Creams \$.0 Check C+ I = = 6. The sum of two numbers is 4, where the difference of the two numbers is 9. Determine the two numbers. Let the first number be A, and the second B, the two equations formed are: A+ B = 4 and A B = 9 Solving b algebra: A+ B = 4 A B = 9 B = B =. A+ B = 4 A + (. ) = 4 A = 6. Solution is A = 6., B =. Check A B = 6. (. ) = 9 (q) Two cables support a weight of 10N as shown in the diagram below. Centre for Teaching and Learning Academic Practice Academic Skills Page 3

24 T 1 T 10N N The equations below appl to the direction indicated; Verticall: 0. 71T T = 10 Horizontall: 0. 71T 1 = + 0.T Calculate the tensions T 1 and T Solving b Determinants 0. 71T T = T 0.T = A = A* A* * = 0. = det A* T 1 = = = = det A det A** T = = = = det A The solution is T = N T = 80. 1N 1 (r) A volume of 6% solution of weedkiller is mied with a different volume of 1% solution to obtain 00mL of a % solution. Calculate the volumes of each of the original solutions. Let the volume of the 6% solution be, and the volume of 1% solution be. This gives an equation: + = 00 Consider the amount of weedkiller in each solution: = = 0 Solving b Algebra: = = = = = 8 = = = 00 = ( ) Solution is.,. Check = = 0 Centre for Teaching and Learning Academic Practice Academic Skills Page 4

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