8 x y 2 x 5 1 Solution: y 6 (1) x 8 y 2 x 5 1 - PDF document

SET 2 Chapter 6 Linear Equations and Inequalities in Two Variables لانيريغتمب ةيطخلا تانيابتملا و تلبداعم 6.1 Linear Equations in Two Variables نيريغتمب ةيطخلا تلبداعملا(لأا تلبداعملاةينً نيريغتمب)  Two linear equations in two variables are solved simultaneously.  Therefore, they are called two simultaneous )ةـينًآ( linear equations in two variables, or a system of two linear equations in two variables.  Simultaneous equations are solved: (i) By substitution ضيوعتلاب (ii) By elimination فذـحلاب (iii) Graphically ينًايبلا مسرلاب (iv) By matrices and determinants ادذحملا و تافوفصملابت Chapter 6: Linear Equations and Inequalities in Two Variables 1

6 .2 Solution of Two Simultaneous Linear Equations by Substitution ضـيوعتلاب نيريغتمب نيـتيطخ نـيتـينًآ نيتلداعم لح Example 1. Solve the following simultaneous equations by substitution and check your results:  y 6  8 x  y   2 x 5 1 Solution:  y 6  (1) x 8  y   2 x 5 1 (2) From equation (1):  y 6  x 8  8  x 6 y Substitute (8  6 y ) for x in equation (2):  y   (2) 2 x 5 1     2 ( 8 6 y ) 5 y 1     16 12 y 5 y 1     17 y 1 16    17 y 17  y 1 By substituting 1 for y in equation (1) or equation (2), the value of x may be found. Use equation (1) to find x :  y 6  x 8 (1)   x 6 ( 1 ) 8   x 6 8   x 8 6  x 2 Checking the answer by substituting 2 for x and 1 for y in equation (2) gives:        LHS 2 ( 2 ) 5 ( 1 ) 4 5 1 RHS Thus, the solution is x = 2 and y = 1 . Chapter 6: Linear Equations and Inequalities in Two Variables 2

Example 2. Use the substitution method to solve the following simultaneous equations and verify the results:     1 15 r 7 t  t  4 2 r Solution:     15 7 1 (1) r t  t  2 r 4 (2) From equation (2):  t  (2) 2 r 4  4  t 2 r Substituting (4  2 r ) for t in equation (1) gives:     15 7 1 (1) r t      15 r 7 ( 4 2 r ) 1      15 r 28 14 r 1     29 r 1 28    29 r 29  r 1 Substitute 1 for r in equation (2) to find t :  t  2 r 4 (2)   2 ( 1 ) t 4   2 t 4   4 2 t  t 2 Check the solution by substituting r = 1 and t = 2 in equation (1):     15 r 7 t 1 (1) ?     15 ( 1 ) 7 ( 2 ) 1 ?     15 14 1    Yes  The solution is r = 1 and t = 2 . 1 1 Chapter 6: Linear Equations and Inequalities in Two Variables 3

Example 3. Solve the following simultaneous equations by substitution and check the results:  y  2 x 3 17  y  4 x 5 1 Solution:  y  2 x 3 17 (1)  y  4 x 5 1 (2) From equation (1):   2 x 3 y 17   2 x 3 y 17  3 y 17  x 2 3  y 17 Substitute for x in equation (2): 2  y  (2) 4 x 5 1    3 y 17     4 5 1 y   2 Multiply both sides by 2:    4 ( 3 y 17 ) 10 y 2    12 y 68 10 y 2   22 y 2 68   22 y 66  66  y 22   y 3 Substitute y =  3 in any of the two equations to find x . Use equation (1) to find x :  y  (1) 2 x 3 17    2 x 3 ( 3 ) 17   2 x 9 17   2 x 17 9  2 8 x 8  x 2  x 4 Verify the results by substituting 4 for x and  3 for y in equation (2):        LHS 4 ( 4 ) 5 ( 3 ) 16 15 1 RHS Hence, the solution is x = 4 and y =  3 . Chapter 6: Linear Equations and Inequalities in Two Variables 4

6.3 Solution of Two Simultaneous Linear Equations by Elimination لح نيتلداعم نـيتـينًآ نيـتيطخ نيريغتمب لابفذـح Example 4 . Solve the following simultaneous equations by elimination and verify the results:  y  2 3 8 x  y 2  x 3 Solution:  y  2 x 3 8 (1)  y 2  (2) x 3 To eliminate x , multiply equation (2) by 2 first:  y 2  (2) x 3  y  2 ( x 2 ) 2 ( 3 )  y  2 x 4 6 (3) Then subtract equation (3) from equation (1):  y  (1) 2 x 3 8  y  (3) 2 x 4 6   Subtract 0 y 2 Thus y =  2 To find x , substitute y =  2 in equation (1):  y  (1) 2 x 3 8    2 x 3 ( 2 ) 8   2 x 6 8   2 x 8 6  2 x 14 14  x 2  x 7 Check the solution by substituting 7 and  2 for x and y respectively in equation (2):  y 2  (2) x 3 ?    ( 7 ) 2 ( 2 ) 3 ?   7 4 3  Yes 3 3 Therefore, the solution is x = 7 and y =  2 . Chapter 6: Linear Equations and Inequalities in Two Variables 5

Example 5. Solve the following simultaneous equations by elimination and verify the results:  b   3 a 4 22  b   7 a 3 2 Solution:  b   3 a 4 22 (1)  b   (2) 7 a 3 2 Multiply equation (1) by 3:  b   (1) 3 a 4 22  b   3 ( 3 a 4 ) 3 ( 22 )    9 a 12 b 66 (3) Then multiply equation (2) by 4:  b   7 a 3 2 (2)  b   4 ( 7 a 3 ) 4 ( 2 )    28 a 12 b 8 (4) To eliminate b , add equations (3) and (4):    9 a 12 b 66 (3)    28 a 12 b 8 (4) 37  0   Add a 74  74    Thus a 2 37 Substituting  2 for a in equation (1) gives:  b   3 a 4 22 (1)     3 ( 2 ) 4 b 22     6 4 b 22     4 22 6 b    4 b 16  16   b 4  4 Check the results by substituting a =  2 and b = 4 in equation (2) :  b   7 a 3 2 (2) ?     7 ( 2 ) 3 ( 4 ) 2 ?     14 12 2    Yes  The solution of the system is a =  2 and b = 4 . 2 2 Chapter 6: Linear Equations and Inequalities in Two Variables 6

6.4 Solution of Two Simultaneous Linear Equations Graphically لح نيتلداعم نـيتـينًآ نيـتيطخ نيريغتمب ابينًاـيبلا مـسرل To solve a system of two linear equations in two variables graphically:  Draw the graphs of the two equations on the same graph.  The graphs are two straight lines that intersect at a specific point.  The point of intersection of the two lines represents the solution of the system. Example 6 . Solve the following simultaneous equations graphically and verify the results:  y  2 x 5  y  2 x 3 Solution: Draw equation (1) by finding the x -intercept and the y -intercept: x -intercept  y  (1) 2 x 5   2 x ( 0 ) 5  x 2 . 5 y -intercept  y  (1) 2 x 5   2 ( 0 ) y 5   y 5 Use (2.5, 0) and (0,  5) to draw equation (1). Similarly, use the x -intercept and the y -intercept to graph equation (2): x -intercept  y  (2) 2 x 3   2 x ( 0 ) 3  x 1 . 5 y -intercept  y  (2) 2 x 3   2 ( 0 ) y 3  y 3 Use (1.5, 0) and (0, 3) to draw equation (2). Chapter 6: Linear Equations and Inequalities in Two Variables 7

The graph of the two equations is shown in the following figure. y 5 4 3 2 1 x 1 2 3 4 5 6 – 1 Solution (2,  1) – 2 – 3 – 4 – 5    From the graph, and x 2 y 1 Verify the results:  y  (1) 2 x 5        LHS 2 ( 2 ) ( 1 ) 4 1 5 RHS  y  (2) 2 x 3        LHS 2 ( 2 ) ( 1 ) 4 1 3 RHS Hence, the solution of the system is x = 2 and y =  1. Chapter 6: Linear Equations and Inequalities in Two Variables 8

6.5 Linear Inequalities in Two Variables تانيابتملا لايطخ ةنيريغتمب  The solution of linear inequalities in two variables x and y is all points ( x , y ) that satisfy the inequality. Example 7 . Find the solution for  y  graphically. 4 x 3 12 Solution:  y  4 x 3 12 : 1- Draw the graph of x - intercept  y  4 x 3 12   4 x 3 ( 0 ) 12  x 3 y - intercept  y  4 x 3 12   4 ( 0 ) 3 y 12   y 4  y  Thus, use (3, 0) and (0,  4) to draw the equation 4 x 3 12 as a solid line. 2- Choose a random point other than y the boundary line points to test. Test (1, 1): 5 4 ? 3 ? 2 Yes  The region above the 1 (1, 1) boundary line is the x solution region. 1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 Chapter 6: Linear Equations and Inequalities in Two Variables 9