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

Chapter 6: Linear Equations and Inequalities in Two Variables

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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

SET 2

Chapter 6

Linear Equations and Inequalities

in Two Variables لانيريغتمب ةيطخلا تانيابتملا و تلبداعم

2

Chapter 6: Linear Equations and Inequalities in Two Variables

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:

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6   y x 1 5 2    y x Solution:

8 6   y x

(1)

1 5 2    y x

(2)

From equation (1): 8 6   y x y x 6 8  Substitute (8  6y) for x in equation (2):

1 5 2    y x

(2)

1 17 17 16 1 17 1 5 12 16 1 5 ) 6 8 ( 2                 y y y y y y y

By substituting 1 for y in equation (1) or equation (2), the value of x may be found. Use equation (1) to find x : 8 6   y x (1)

2 6 8 8 6 8 ) 1 ( 6        x x x x

Checking the answer by substituting 2 for x and 1 for y in equation (2) gives: RHS 1 5 4 ) 1 ( 5 ) 2 ( 2 LHS        Thus, the solution is x = 2 and y = 1.

Chapter 6: Linear Equations and Inequalities in Two Variables

3 Example 2. Use the substitution method to solve the following

simultaneous equations and verify the results:

1

7 15     t r

4

2  t r Solution:

1 7 15     t r (1) 4 2  t r

(2)

From equation (2):

4 2  t r

(2)

r t 2 4 

Substituting (4  2r) for t in equation (1) gives: 1 7 15     t r (1)

1 29 29 28 1 29 1 14 28 15 1 ) 2 4 ( 7 15                   r r r r r r r

Substitute 1 for r in equation (2) to find t : 4 2  t r (2)

2 2 4 4 2 4 ) 1 ( 2        t t t t

Check the solution by substituting r = 1 and t = 2 in equation (1): 1 7 15     t r (1)

1 1 1 14 15 1 ) 2 ( 7 ) 1 ( 15            ? ? Yes  The solution is r = 1 and t = 2.

4

Chapter 6: Linear Equations and Inequalities in Two Variables

Example 3. Solve the following simultaneous equations by substitution and check the results: 17 3 2   y x 1 5 4   y x Solution:

17 3 2   y x (1) 1 5 4   y x (2) From equation (1):

2 17 3 17 3 2 17 3 2       y x y x y x

Substitute

2 17 3  y

for x in equation (2):

1 5 4   y x

(2)

1 5 2 17 3 4          y y

Multiply both sides by 2:

66 22 68 2 22 2 10 68 12 2 10 ) 17 3 ( 4           y y y y y y 3 22 66     y y

Substitute y = 3 in any of the two equations to find x. Use equation (1) to find x :

17 3 2   y x

(1)

4 2 8 8 2 9 17 2 17 9 2 17 ) 3 ( 3 2           x x x x x x

Verify the results by substituting 4 for x and 3 for y in equation (2): RHS 1 15 16 ) 3 ( 5 ) 4 ( 4 LHS        Hence, the solution is x = 4 and y = 3.

Chapter 6: Linear Equations and Inequalities in Two Variables

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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: 8 3 2   y x 3 2   y x Solution:

8 3 2   y x (1)

3 2   y x

(2) To eliminate x, multiply equation (2) by 2 first:

3 2   y x

(2)

) 3 ( 2 ) 2 ( 2   y x

6 4 2   y x (3) Then subtract equation (3) from equation (1):

8 3 2   y x

(1) 6 4 2   y x

(3)

Subtract 0

y  2 

Thus y = 2 To find x, substitute y = 2 in equation (1):

8 3 2   y x

(1)

8 6 2 8 ) 2 ( 3 2      x x 7 2 14 14 2 6 8 2      x x x x

Check the solution by substituting 7 and  2 for x and y respectively in equation (2):

3 2   y x

(2)

3 3 3 4 7 3 ) 2 ( 2 ) 7 (      

Therefore, the solution is x = 7 and y = 2.

? ? Yes

6

Chapter 6: Linear Equations and Inequalities in Two Variables

Example 5. Solve the following simultaneous equations by elimination and verify the results:

22 4 3    b a 2 3 7    b a

Solution:

22 4 3    b a

(1)

2 3 7    b a

(2) Multiply equation (1) by 3:

22 4 3    b a

(1)

) 22 ( 3 ) 4 3 ( 3    b a

66 12 9    b a (3) Then multiply equation (2) by 4:

2 3 7    b a

(2)

) 2 ( 4 ) 3 7 ( 4    b a 8 12 28    b a

(4) To eliminate b, add equations (3) and (4): 66 12 9    b a (3) 8 12 28    b a (4) Add

a 37  0 74  

Thus

2 37 74     a

Substituting 2 for a in equation (1) gives:

22 4 3    b a

(1)

4 4 16 16 4 6 22 4 22 4 6 22 4 ) 2 ( 3                    b b b b b

Check the results by substituting a = 2 and b = 4 in equation (2):

2 3 7    b a

(2)

2 2 2 12 14 2 ) 4 ( 3 ) 2 ( 7            Yes  The solution of the system is a =  2 and b = 4. ? ?

Chapter 6: Linear Equations and Inequalities in Two Variables

7

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

• f the system.

Example 6. Solve the following simultaneous equations graphically and verify the results: 5 2   y x 3 2   y x Solution:

Draw equation (1) by finding the x-intercept and the y-intercept: x-intercept

5 2   y x

(1)

5 . 2 5 ) ( 2    x x

y-intercept

5 2   y x

(1)

5 5 ) ( 2     y y

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

3 2   y x

(2)

5 . 1 3 ) ( 2    x x

y-intercept

3 2   y x

(2)

3 3 ) ( 2    y y

Use (1.5, 0) and (0, 3) to draw equation (2).

8

Chapter 6: Linear Equations and Inequalities in Two Variables

The graph of the two equations is shown in the following figure. From the graph,

2  x

and

1   y

Verify the results:

5 2   y x

(1) RHS 5 1 4 ) 1 ( ) 2 ( 2 LHS       

3 2   y x

(2)

RHS 3 1 4 ) 1 ( ) 2 ( 2 LHS        Hence, the solution of the system is x = 2 and y = 1.

y

1 2 3 4 5 6 2 3 4 1

–1 –2 –5

x

Solution (2, 1)

–3 –4

5

Chapter 6: Linear Equations and Inequalities in Two Variables

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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

12 3 4   y x graphically.

Solution:

1- Draw the graph of

12 3 4   y x

:

x - intercept 12 3 4   y x

3 12 ) ( 3 4    x x

y - intercept 12 3 4   y x

4 12 3 ) ( 4     y y

Thus, use (3, 0) and (0, 4) to draw the equation 12 3 4   y x as a solid line.

2- Choose a random point other than

the boundary line points to test. Test (1, 1):

y

1 1 2 3 4 5 6 2 3 4

–1 –2 –5

x

(1, 1)

–3 –4

5

? ? Yes  The region above the boundary line is the solution region.

10

Chapter 6: Linear Equations and Inequalities in Two Variables 3- Shade the solution region as in the figure below.

Example 8. Graph the solution to the following system of inequalities:

12 2   y x 6 3   y x

Solution:

To graph the solution to a system of inequalities: 1- Graph each inequality on the same graph. 2- See where the shading of the inequalities overlaps. The

• verlapping region is the solution region to the system
• f inequalities.

For

12 2   y x

, use (12 , 0) and (0, 6) to draw the boundary line, and then test point (0, 0):

12 12 ) ( 2 ) ( 12 2      y x

For

6 3   y x

, use (2 , 0) and (0, 6) to draw the boundary line, and then test point (0, 0):

6 6 ) ( ) ( 3 6 3      y x

Yes  The region below the boundary line is the solution region. ? ? No  The region below the boundary line is the solution region.

Chapter 6: Linear Equations and Inequalities in Two Variables

11 The solution to the system is the region that contains the shading

• f the two inequalities and as shown in the figures below.