Mathematics

Key unit competence

By the end of this unit, I will be able to solve problems related to simultenous linear equations, inequalities and represent the solution graphically. Let us now learn about equations containing two variables

3.1 An equation in two variables

Activity 3.1

1.  On a market day, Jean bought some white chicken and some black ones. In total he bought 12 chicken. Choose letters of your choice to represent the total number of chicken in an equation.

2. Lucy bought some oranges and mangoes. Mary bought twice the number of oranges as Lucy and thrice the number  of mangoes. If Mary bought, 18 fruits in total, represent her total number of fruits in an equation.

Suppose Erick and Robert together have eight children. How can we express this situation in a mathematical statement? Since there are two numbers that we do not know, it is natural to use two variables (unknowns).

Thus, if Erick has x children and Robert has y children, together they have (x + y) children. This mean that   x + y = 8

Table 3.1 below shows possible pairs of numbers which make the above equation true.

In table 3.1 above, for each value of x, there is a corresponding value of y. We say that such a pair of numbers satisfies the equation or it is a solution of the equation.

Activity 3.2

At a certain point, two small businessmen noted that their bank accounts had 110 and 600 dollars respectively. The first man decided to increase his account by 30 dollars every year for n years. th second decreased his bank account balance by 40 dollars every year for n years.

After n years, the bank balance in each year was p dollars. Assuming that no other transactions were done in these accounts,

(i) Express p in terms of n in each case

(ii) Use each equation to make a table of values for n and p for each person, for values of n not more than 10 years.

(iii) Do you think there will be a time when their bank balances will be equal? If yes, when and how much will it be?

(iv) How do you describe the relations in (ii) above.

Observations

From this activity you should have observed that:

(i) The two situations can be represented by the equations

p = 30n +110 and p = 600 – 40n.

(i) Tables 3.2 and 3.3 below show the required tables for p =110 + 30n, see Table 3.2.

After 7 years, both account balances will be 320 dollars. The equations p = 30n + 110 and  p = 600 – 40n are called simultaneous equations because there are two distinct variables n and p.

The solution set of the equations are n = 7 and p = 320 as seen from the table

Simultaneous equations are a set of equations with same set of two or more variables that collectively satisfy all the equations.

Example 3.1

Show that the following pairs of numbers satisfy the equation x + 3y = 18.

(a)  (0, 6)      (b)  (3, 5)   (c)  (–3, 7)     (d)  (21, –1)

Solution

In these pairs of numbers, the first number represents the value of x, while the second one represents the value of y.

(a)  x + 3y   =  18 : (0, 6)         

LHS   =  0 + 3 × 6  

            =  0 + 18    

               = 18       

∴ LHS  =  RHS 

(b)  x + 3y   =  18 : (3, 5)   

         LHS   =  3 + 3 × 5              

                    =  3 + 15     

                     =  18 

∴LHS  =  RHS   (c)  x + 3y =  18 : (–3, 7)    

   LHS   =  –3 + 3 × 7              

              =  –3 + 21     

                =  18 

∴LHS  =  RHS (d)  x + 3y   =  18 : (21,–1)  

  LHS   =  21 + (3 × –1)              

             =  21 + –3     

            =  18 

∴LHS  =  RHS

In each case, the left hand side of the equation is equal to the right hand side; all the given pairs of numbers satisfy the equation.

Exercise 3.1

In Questions 1 and 2, the given pairs of numbers are such that the first number represents the value of x, while the second represents the value of y.

1. Show that the given pairs of numbers satisfy the equation y + 2x = 12.

(a) (1, 10)                          (b)  (5, 2) 

(c) (0, 12)                          (d)  (8, –4)

2. Which of the following pairs of numbers satisfy the equation 3x + 4y = 7?

(a) (1, 0)                             (b)  (1, 1) 

(c)  (5, –2)                         (d) (2, 2)  

(e)  (7, 3)                          (f)  (9, –5)

3. If x and y represent whole positive numbers, give the first four pairs of numbers which satisfy the equation  3x + y = 15. 4. Copy and complete table 3.4 below for pairs of  numbers that satisfy the equation x + 3y = 17.

5. If x and y are restricted to positive whole numbers, give six pairs of numbers that satisfy each of the following equations:

(a) 3x –  y  =  8

(b) x – 2y  =  1

Is there any pair of numbers that satisfies  both equations? If yes, state the pair.

3.2 Solving simultaneous equations

3.2.1 Graphical solutions of simultaneous equations

Activity 3.3

1.  Draw on the same axes the groups of the equations.  

x + y = 4 ...... (i)  

2x + y = 5 ........ (ii)

2.  Read the x and y coordinates of the point of intersection of the two lines.

3. Substitute these x and y values in each of the equation and test whether they satisfy the equations or not. What do you notice.

4. What can you say about the coordinates of intersection of two graphs in relation to the solutions of the two equalities.

From Activity 3.3, we have seen that linear graphs can be used to determine the values of the variables that satisfy a system of equations. Simultaneous linear equations in two unknowns may be solved by plotting the two straight lines then note the coordinates of the point of intersection. If for example, the coordinates of the point of intersection are (2, 3), then we say the solution of the linear simultaneous equations is  x = 2, y = 3.

Consider the pair of simultaneous equations

y + 2x = 6  

x – y = 3

Table 3.5 (a) below gives some of the ordered pairs of values that satisfy the equation
 y + 2x = 6.

Table 3.5 (b) gives pairs of values for the equation x – y = 3.  

The ordered pair x = 3, y = 0, i.e. (3, 0) appears in both tables; and it is the only pair that does so. It is the only pair of values that satisfies both equations simultaneously (i.e. at the same time). Hence, the solution of the simultaneous equations is x = 3, y = 0.

The result can be obtained by drawing the graphs of the two lines as in Fig. 3.1. The two lines intersect at the point (3, 0). This is the only point that is on both lines.
 

The coordinates of the point at which the lines intersect give the solution of the simultaneous equations they represent.

Example 3.2

Solve graphically the simultaneous equations  

x – 2y = –1  2x – y = 4

Solution

Step 1:

Make a table of values for each equation. Three pairs of values are sufficient for each (Table 3.6 (a) and (b)).

Step 2:

Choose a suitable scale and plot the points. Draw the lines. Extend  if necessary, so that they  intersect. Fig. 3.2 shows the graphs of the two lines.

Step 3:

Read the coordinates of the point of intersection. From the graph, the lines intersect at (3, 2). The solution of the simultaneous equations is x = 3,  y = 2

Exercise 3.2 Solve graphically the following pairs of  simultaneous equations.

1. x + y = 3                   2.  2x = 3 + y

y = 3x – 1                     7x + 2y = 16

3. 2x – y = 3              4. 3x + 2y = 0

2y = –x + 14                 5x + y = 7

5. y + 1 = 2x              6. 3y – x = 4

2y + x + 7 = 0             2x – 5y = –7

7. 3x + 4y = 3.5           8. 2y + 3x = –5

7x – 6y = 0.5                   3y + 2 = x

9. 4x – y = 2                  10. 4x – 2y = 4

6x + 4y = 25                   2x – 3y = 0

11. 2x – y = –1              12. 12x + 6y = 12

  x – 2y = 4                    2x – 3y = –2

13. 10x – l0y = 3            14. x – y = –1

2x + 3y = 3.1                     4x – 8y = 4

15. 2y + 2 = 3x – 6                               16. 5x – 2y = 1            

(y – 1) /2 +( x – 3)/ 2 = (x + 3)/ 3            4x + 3y = –10.7

Classification of simultaneous equations

Activity 3.4

Using a graph paper, draw on separate graphs the pairs of lines whose equations are  

(a)  2x – y  = –1 

x – 2y = 4

(b)  x – 2y = –1  

2x – 4y  = –2

(c) 2x – y  = –4 

4x – 2y = 6

In each case describe completely the resulting graphs

Observation

Fig 3.3 shows the graphs of lines whose equations are x – 2y = 0 and 2x + y = 5 The two lines intersect at point (2, 1). thus x = 2 and y = 1 satisfy the two equations simultaneously. Therefore, we say there is one unique solution to the two equations.

Such a pair of equations is classified as consistent and independent. Fig 3.4 shows the graph of the lines whose equations are x – 3y = –1 and 2x – 6y = –2.

The lines representing the two equations are coincident i.e. all the points on one line lie on the second line. We say that there is an infinite number of solutions to the two equations you should have observed that one equation is a multiple of the other. Such equations are classified as consistent and dependent.

Fig 3.5 above shows the graphs of x – y = 4 and x – y = 2. The lines do not touch and therefore no value of x or y can satisfy both the equations. They have no solutions and therefore are classified as inconsistent/ incompatible.

Fig 3.5 above shows the graphs of x – y = 4 and x – y = 2. The lines do not touch and therefore no value of x or y can satisfy both the equations. They have no solutions and therefore are classified as inconsistent/ incompatible.

Note that the lines are parallel and therefore will never meet. Therefore, no pair of values of x and y can satisfy both equations.

Summary

(i) Two or more simultaneous equations have a unique solution if the lines representing them intersect.

(ii) Two equations have an infinite solution if their lines are coincident.

(iii) Equations have no solutions if the lines representing them are paralle

Exercise 3.3

For each question in this exercise, draw on the same axis and classify the sets of equations. In each case state the values of x and y that satisfy the equations simultaneously, where possible.

1. 3x + 2y = 6 and x – y = 2

2. 2x – 3y = 6 and 2x – 3y =  –15

3. 2x – y = 2 and x + y = 4

4. x + 2y = 6 and x + 2y = 8

5. 3x – 2y = 5 and 2x - y =  –1

6. 2x – 3y = 5 and 2x - y =  –1

7. x – 3y = 4 and x – 3y = 8

8. 2x – y =  –3 and x = y = 0

9. x – y =  + 8 and  –3x + 3y + 24 = 0

10. y – 2x =  –3 and  –2y + 4x = 6

3.2.2 Solving simultaneous equations analytically

From the foregoing work, we have seen that simultaneous equations can be solved graphically. However, for some equations it is not possible to obtain a totally accurate graph so that the reading at the point of intersection is not accurate either. Therefore, we use alternative algebraic methods of solving simultaneous equations.

These methods include:

i. Substitution method

ii. Elimination method

iii. Comparison method and

iv. The Cramer’s rule

3.2.2.1 Solving simultaneous equations by substitution method

Activity 3.5

Consider the equations: 

2x + y  = 7 ....... (i)

3x – 2y = 0 ....... (ii)

Using equation (i), express y interms of x. In equation form and label this equation (iii) i.e

y = ____ ... (iii)

1. Substitute the value of y (in terms of x) from equation (iii) into equation (ii) to have equation (iii) in terms of x only.

2. Solve equation (iii) to get the exact value of x.

3. Substitute the exact value of x in equation (i) or (iii) to get the value of y.

4. Confirm whether the values of x and y satisfy both equations (i) and (ii)

5. What value is given to this method of solving simultaneous equations?

Consider the following equations

3x – 5y = 23.............(1) 

x – 4y = 3..............(2)

Using equation (2), add 4y to both sides:   

x + 4y = 3  x – 4y + 4y = 3 + 4y     

x  = 3 + 4y..............(3)

In equation (3), x is said to be expressed or solved in terms of y. To express the equation 3x – 5y =  23 as an equation in one variable, substitute
 (3 + 4y) in place of x in equation (1):

 3x – 5y = 23 becomes   

3(3 + 4y) – 5y = 23    

9 + 12y – 5y = 23             

9 + 7y = 23                

7y = 14                       

y = 2

Substituting y = 2 in equation

(3) we get;          

x  = 3 + 4(2)  

= 3 + 8     

∴        x  = 11

Use equation (1) to check that the solutions are correct. This method of solving simultaneous equations is called substitution method.

Solution

Multiply (1) by 12 to remove the denominators.
 4x + 4y – 3x + 3y = 8   

x + 7y = 8..............

(3) Multiply (2) by 12 to remove the denominators.
 8x – 12 – 6y – 9 = –19    

8x – 6y = 2..............

(4)  Using (3), express x in terms of y:      x = 8 – 7y..............

(5) Substitute (5) in (4) to eliminate x:

8(8 – 7y) – 6y =  2  64 – 56y – 6y =  2    64 – 62y =  2      

   62y =  62       

  y =  1
 Substitute y = 1 in (5):  

x = 8 – 7(1)      

= 8 – 7      

x = 1

Substitute y = 1 and x = 1 in equation (1) or (2) to check that the solutions are correct.
Solution is s = {(1,1)}

To solve simultaneous equations by substitution method:

1. First decide which variable is easier to substitute.

2. Using the simpler of the two equations,  express the variable to be eliminated in terms of the other i.e. make it the subject of the formulae.

3. Using the other equation, substitute the equivalent for the variable to be  removed.

4. Solve for the remaining variable.

5. By substitution, solve for the other unknown.

6. By substitution, check whether your solutions satisfy the equations.

Exercise 3.4

In questions 1 to 4, find y in terms of x.

1. 4x – y = 12                        2.  2x + 5y = 10

3.  1 3 x– 4y = 16                    4.  3x – 1 4 y = 10

In questions 5 to 8, express x in terms of y.

5. x – 5y = 3                          6.  9x + 4y = 0

7. 1/3 x = 1/6 (y – 1)               8.  1/3 (x – 2) – y = 2

Use substitution method to solve the following pairs of simultaneous equations. 9. a + b = 3                              10.  w – 2z = 5

4a – 3b = 5                               2w + z = 5

11. x + y = 0                             12.  x = 5 – 2y

2y – 3x = 10                               5x + 2y = 1

13. 4x – 3y = 1                      4.  6a – b = –1

x – 4 = 2y                                  4a + 2b = –6                     

15. 4m – n = –3                      16.  5q + 2p = 10

8m + 3n = 4                               3p + 7q = 29

17. 1/v + 1 u = 1/5                     18.  2s – 4t = 8 1

v – 1 u = 2 5                                3s – 2t = 8

19. 2x – 4y + 10 = 0                20. 1/2 a – 2b = 5

3x + y – 6 = 0                                  1/2 a + b = 1

21. 2y/5 + z/3 = 2 (2/3)                   22.  (a – 1) /2 + (b + 1)/5 = 1/5

y = 2(z + 1)                                              a + b 3 =  b – 1

3.2.2.2 Solution of simultaneous equations by elimination method

Sometimes the substitution method gives rise to awkward fractions when attempting to express one variable in terms of the other. In such a situation it is advisable to use the method of addition or subtraction of the given equations. This method is also called the elimination method.

Activity 3.6

Consider the equations

3x - 2y = 11………………..(1)

5x + 2y = 29………………..(2) (

i)  Add the left hand sides of equations (1) and (2).

(ii) Add the right hand sides of equations (1) and (2).

(iii) Equate the results of (i) and (ii) above to obtain equation (3). What do you notice?

(vi) Solve equation (3).

(v) By substitution use the solution of equation (3) to obtain equation (4).

(vi) Solve equation (4).

Summaries your findings

From activity 3.6 above you should have observed the following: Since left hand side of each equation equals its right hand side, adding left hand sides of (1) and (2) equals the result of adding the corresponding right hand side

i.e. 3x – 2y = 11

 5x + 2y = 29

8x + 0 = 40

In doing this addition, the term in y disappears leaving a simple equation in x Thus 8x = 40 x

               =  40/8

                = 5

Substituting 5 for x in equation 1,   

3x – 2y = 11 becomes 3(5) – 2y = 11   

15 – 2y = 11  

–2y = –4       

y = –4/ –2

= 2

The solution is x = 5, y=2 This process of getting rid of one of the variables is called elimination method. Now consider the equations

2x - 5y = 27……………(1)

2x + 3y = 3……………..(2)

Follow steps similar to the ones used in the above activity to eliminate one of the variables and summarise your findings.

Observations

In the equations, we can eliminate the terms in x by subtraction.

2x – 5y  =  27……………….(1)

2x + 3y =  3………………….(2)

-8y = 24

By substraction, the term in x disappear so  

–8y = 24
 y =  24/–8

y =  –3

Substituting  -3 for y in equation

(2) 2x + 3y  =

3 becomes 2x + 3(–3)  = 3      

2x – 9 = 3

 x =  12/2 =  6

The solution is x = 6, y = –3

To use elimination method,

1. Decide which unknown is easier to eliminate.

2. Solve for the remaining unknown.

3. When one unknown has been found, obtain the other by substituting in the easier of the original equations.

Example 3.4

Solve the simultaneous equations 

3x + 2y = 12.............(i) 

4x – 2y = 2...............(ii)

Solution

If we add equation (i) and (ii), we get a simpler equation in one unknown. We do this by adding LHS of equation (i) to the LHS of equation (ii) and the RHS of equation (i) to RHS of equation (ii).

Thus,

In any of the original equations, we can use 2 instead of x, i.e. we can substitute 2 for x to get the value of y.

Using equation (1), 3x + 2y = 12 becomes 

3 × 2 + 2y  = 12              

6 + 2y  = 12              

2y  = 6               

y  = 3

The solutions of the simultaneous equations are therefore x = 2 and y = 3 or Solution is s = {(2,3)}

Example 3.5 Solve the simultaneous equations 

2x + 4y = –12 ......... (i)

5x + 4y = –33 ......... (ii)

Solution

We do this by subtracting LHS of equation (ii) from LHS of equation (i) and the RHS of equation 2 from RHS of equation (i). If we subtract equation (ii) from equation (i) we get a simple equation in one unknown.

  x  =  –7 To make the given equations true, x must be equal to –7. Now, in equation (i) or (ii) we use –7 instead of x to solve for y.

Thus,

2x + 4y = –12 becomes   

(2 × –7) + 4y = –12

i.e.         –14 + 4y = –12                  

                    4y  =  2             

                   y  = 1/2

Use –7 for x in equation (2) and confirm

that y = 1/2 .

Solution is s = {(–7,1 2 )}

In Examples 3.4 and 3.5 we were able to get rid of one of the variables by addition or subtraction, because the coefficients of one of the unknowns in both equations had either the same sign or opposite sign. You should note that if the coefficients of the variable to be eliminated have the same sign, we subtract the equations, otherwise add the equations. This method of getting rid of one of the variables by addition or subtraction is known as the elimination method.

Exercise 3.5

In this exercise, examine each of the pairs of equations carefully, and decide when to add and when to subtract. Then solve the simultaneous equations.

 1. 3x – y = 8                         2.   5x – y = 18

     x + y = 4                                  3x + y = 14

3. 3x – 2y = 0                           4. 4x – 3y = 16

x – 2y = –4                                2x + 3y = 26

5. x + 2y = 11                             6. 3x + 2y = 12

x – 2y = 3                                        4x + 2y = 2

7. 3a – 2b = 11                                8. r – s = 1

2a – 2b = 10                                        –r – s = 13

9. 2m – n = 11                                 10. 5x + 3y = 9

3m + n = 49                                         –5x + 2y = 1

11. 7x – 2y = 29                                  12. 10x – 3y = 36

7x + y = 38                                                 x + 3y = 18

13. 5x – 6y = 16                                     14. x + 6y = –5

7x + 6y = 44                                             x – 9y = 0

15. 6x + 4y = 24                                        16. 5x + 3y = 77

7x – 4y = 2                                                       15x – 3y = 3

Solving complex simultaneous equations by elimination method

Sometimes it may not be obvious which unknown to eliminate or how to eliminate. Consider the equations:

3x – 2y = 8

x + 5y = –3

In this pair of equations, we cannot eliminate any variable by simple addition or subtraction. We must first make the coefficient of  x or y the same in both cases. Then we may be able to add or subtract as before. The examples that follow illustrate this process

Example 3.6

Use elimination method to solve the simultaneous equations 

3x – 2y = 8  

x + 5y = –3

Solution

In this case we choose to eliminate x.
 3x – 2y = 8.............(i)
 x + 5y = –3.............(ii)

Leave (i) as it is:      3x – 2y =  8 ....(i)  

Multiply (ii) by 3:     3x + 15y =  –9 ...(iii)

Subtract (iii) from  (i) –17y  =  17                  

y   =  –1 Use –1 instead of y in (i)  

3x – 2 ( –1) = 8        

3x + 2 = 8               

3x = 6                 

x = 2 Check in equation (i):

LHS = 2 + 5 (–1) = 2 – 5 = –3,

RHS = –3

Example 3.7

Solve the simultaneous equations
 3x + 4y = 10.............(i)
 2x – 3y = 1...............(ii)

Solution

Let us eliminate y in this case.

Multiply (i) by 3:    9x + 12y = 30  ...(iii)   

Multiply (ii) by 4:    + 8x – 12y = 4  ...(iv) 

  Add (iii) to (iv):      17x      

= 34              

x = 2     

Substitute x = 2 in (i): 3 × 2 + 4y = 10               

6 + 4y = 10                     

4y = 4                         

y = 1

Check in equation

(ii): LHS = (2 × 2) – (3 × 1) = 4 – 3 = 1, 

RHS = 1

Solution is s = {(2,1)}

Example 3.8

Solve the simultaneous equations  

2x + 7y = 15  5x – 3y = 19

Solution

2x + 7y = 15...........(i) 5x – 3y = 19...........(ii)

Eliminate x: (i)  × 5: 10x + 35y = 75...........(iii)        

                     (ii)  × 2: 10x – 6y = 38 ...........(iv)

                      (iii) – (iv):   41y = 37         

                            y  = 37/41

Note that since y is a fraction, substituting y = 37 41 in equation (i) or (ii) would make work more difficult. We obtain x by eliminating y (as was done for x)

i.e.
 (i)  ×  3: 6x + 21y  = 45.............(v)

(ii)  ×  7: 35x – 21y = 133............(vi)

(iii) + (vi): 41x  = 178       

    ∴x  = 173/41 = 4 (14/41)

Hence, the solution is x = 4 (14/41) ,  y = 37/41

Solution is s = {(4(14/41), 37/41)}    

To solve simultaneous equations by elimination method:

1. Decide which variable to eliminate.

2. Make the coefficients of the variable the same in both equations.

3. Eliminate the variable by addition or subtraction as is appropriate.

4. Solve for the remaining variable.

5. Substitute your value from 4 above in any of the original equations to solve for the other variable.

Exercise 3.6

Solve the simultaneous equations.

1. 3x + 2y = 16                 2. 2x – y = 7

2x – y = 6                           5x – 3y = 16

3. 2x + 3y = 27                  4. 3x + 2y = 13

3x + 2y = 13                           2x + 3y = 12

5. x = 5 – 2y                        6. x + y = 0

5x + 2y = 1                             2y – 3x = 10

7. 3x + y = 12                     8. 4y – x = 7

2x – 3y = 8                             3y + 4x = –9

9. 5n + 2m = 10                   10. 2x – 4y = 8

3m + 7n = 29                             3x – 2y = 8

11. 2x + 3y = 600                 12. 6a – b = –1

x + 2y = 350                                4a + 2b = –6

13. w – 2z = 5                         14. 2x – 4y = –10

2w + z = 5                                      3x + y – 6 = 0

15. 9x + 3y = 4                          16. 2x – y = 6

3x – 6y = –1                                      3x + 2y = 18

17. 3x – 4y = –5                            18. 2x – 7y = –10

2x + y = 6                                        9y + 5x = 6

3.2.2.3 Comparison method

We have just learned to solve simultaneous equations graphically, by substitution and by elimination methods. Another method of solving simultaneous equation is the comparison method.

Activity 3.7

Consider the equations (i) x + y = 5   (i)       

                                      (ii) 2x – y = 4 (ii)

1. Using equation (i) x + y = 5, solve for x in terms of y to obtain equation (iii)

2. Solve for x in terms of y using the equation (ii) 2x – y = 4 to obtain equation (iv)

3. Equate equation (iii) to equation (iv) to obtain equation (v)

4. Now solve equation (v) to obtain the value of y.

5. Using the value of x in equation (i) find the value of x

Consider the equation  x + y = 6    (i)  

                                     x – y = 2   (ii)

We can solve these equations using the substitution method twice as was demonstrated in activity 3.7 above. Using equation (i) find y in terms of x

x + y = 6 …… (i)

y = 6 – x ………(iii) …………  

solve for y in terms of x

x – y = 2 ........(ii) –y = 2 – x

y = x – 2 ………………… (iv)  

solve for y in terms of x

6 – x  = x – 2 ………….   

Comparing the values of y in   terms of x –2x = -8

                                                               x = –8 ÷ - 2 = 4

Using equation (i) substitute 4 for x to find y 4 + y = 6  y = 2

∴Solution set is (x, y) = {(4, 2)}

Note:

The equation x + y = 6 and x – y = 2 can also be solved simultaneously using the approach of intersecting lines.

This method makes the following assumptions

(i) The lines represented by x + y  =  6 and x – y = 2 are distinct and intersect at only one point

(ii) The variables x and y are real numbers.

Let any ordered pair(x, y) lie on the line L1 x + y = 6 be defined as l1: x + y = 6, (x, y) ∈ (Real numbers) In L1, y = 6 – x hence,

the ordered pair (x, y)  = (x, 6 – x)

Similarly let any ordered pair(x, y) that lies on l2 x – y = 2, be defined as l2: x – y = 2, (x, y) ∈ (Real numbers) In this case x – y = 2 

hence the ordered pair (x, y)=(x, x – 2)

At the point of intersection, (x, x – 2)  = (x, 6 – x) ⇒    x – 2 = 6 – x But, x + y = 6   2x = 8     4 + y = 6     x = 4          

y = 2

The solution is s = {(4,2)}

Example 3.9

Use the method of comparison to solve the equation

x – 2y = 9 3y – x =  – 11 

Solution

Let x – 2y  =  9 .............................(i)  

– x + 3y  =   – 11 ........................(ii)

And the variables to compare be From equation (i)

x = 9 + 2y  .........(iii)

Equation (ii) x = 11 + 3y ..............(iv)

Comparing (iii) and (iv)

9 + 2y  =  11 + 3y

2y – 3y  =  11 – 9

  –y = 2  

y =  –2 

Using equation (iii),

x = 9 + 2y

When y  =  – 2,

x = 9 + 2(–2)  = 9 – 4     

x  = 5

(x, y) =  (5, – 2)  ⇒  x = 5, y =   –2

Solution is s = {(5, –2)}

Exercise 3.7

Use the comparison method to solve the simultaneous equations. All variables represent real numbers.

1. Solve for x in terms of y in

(a)  x + 2y  =  6                      (b)  x – 3y  =  4   

(c)  3x – 6y  =  2                      (d)  2x – y  = 6

2. Solve for y in terms of x in:

(a) x + y  = 4                                  (b) 3x – y = 2   

(c)  2y – x – 8 – 0                           (d)  2x = y + 4

3. Use the method of comparison to solve the simultaneous equations.

(a) 2y = 2x – 2             

      2y = 4x – 6

(b) x =  –2y + 3        

x = 3y – 7

(c)  y =  –3 – 2x         

        y = 2x – 1

4. Solve by comparison method

(a) 3x = y + 11  

      y = x – 5

(b) 5x – y =  – 13 

      y – 3x = 9

 (c)  3y + 2 = x   

        x + 2y  =  8 – y

(d)  x + 3y  =  3 

     2y – x – 3  =  0

5. Solve the simultaneous equations:

(a)  2x = 3y + 2 

        2x = 6 + y  

(b)  3 – y = 3x 

      3x = 2y + 3     

(c) 2x + 16 = 4y 

    4y – 3x = 18                     

(d) 2x – 5 = 3y 

      5x – 2y = 18

6. Two lines are given by 2x + y = 5 and 4x – y = 1. Find the coordinates of the point of intersection. Hence state the solutions of the simultaneous equations 2x + y = 5 and 4x – y  =  1

7. Solve the simultaneous equations:

(a) 2y + 3x  =   – 5  

     3y + 2  =  x   

(b) 12x + 6y  =  12 

      2x – 3y  =   – 2

3.2.2.4 Cramer’s rule

Activity 3.8

1.  Research the following from reference books and the internet. • Matrix and the order of matrix. • The notation of representing a matrix • How to work out the determinant of 2 × 2 matrix.

2. Discuss with your partner how to calculate the determinant of the matrix