1. 7 Linear Equations in Two Unknowns
7.1 Linear Equations in Two Unknowns and Their
Graphs
7.2 Solving Simultaneous Linear Equations in Two
Unknowns by the Graphical Method
7.3 Method of Substitution
7.4 Method of Elimination
7.5 Applications of Simultaneous Linear Equations
in Two Unknowns

2. 7.1 Linear Equations in Two Unknowns and Their Graphs
The following shows some other linear equations in two unknowns.
They look like linear equations in one unknown, except they have one more variable.

3. 7.1 Linear Equations in Two Unknowns and Their Graphs
A. Solutions of Linear Equations in Two Unknowns

4. 7.1 Linear Equations in Two Unknowns and Their Graphs
A. Solutions of Linear Equations in Two Unknowns

5. Example 1T Solution: 7 Linear Equations in Two Unknowns
Determine whether the following ordered pairs are solutions of the equation 2x  y  4.
(a) (4, 0) (b) (3, –2)
Solution:
(a)
(b)

6. 7.1 Linear Equations in Two Unknowns and Their Graphs
B. Graphs of Linear Equations in Two Unknowns
y  x  2

7. 7.1 Linear Equations in Two Unknowns and Their Graphs
B. Graphs of Linear Equations in Two Unknowns

8. Example 2T Solution: 7 Linear Equations in Two Unknowns
Does C(3, –5) lie on the graphs of the following equations? Explain your answer.
(a) 7x + 5y = –4 (b) 8x – 3y = 39
Solution:
(a) Substitute (3, –5) into the equation 7x  5y  –4,
L.H.S.  7(3)  5(–5)
 –4
 R.H.S.

9. Example 2T Solution: 7 Linear Equations in Two Unknowns
Does C(3, –5) lie on the graphs of the following equations? Explain your answer.
(a) 7x + 5y = –4 (b) 8x – 3y = 39
Solution:
(b) Substitute (3, –5) into the equation 8x – 3y = 39,
L.H.S.  8(3) – 3(–5)
 39
 R.H.S.

10. 7.1 Linear Equations in Two Unknowns and Their Graphs
B. Graphs of Linear Equations in Two Unknowns

11. 7.1 Linear Equations in Two Unknowns and Their Graphs
B. Graphs of Linear Equations in Two Unknowns

12. Example 3T Solution: 7 Linear Equations in Two Unknowns
The figure shows the graph of 2x – 4y  1.
Find the coordinates of C.
Solution:
From the graph, the y-coordinate of C is –1.
Substitute y  –1 into 2x – 4y  1, we have

13. Example 4T Solution: 7 Linear Equations in Two Unknowns
(a) Draw the graph of the equation 2y  x  2.
(b) Using the graph drawn in (a), answer the following questions.
(i) If P(–3, y) lies on the graph of 2y  x  2, find the value of y.
(ii) Is (3, 2.5) a solution of the equation 2y  x  2?
Solution:
2y  x  2
(a) 2y  x  2 x 2 2 y 1

14. Example 4T Solution: 7 Linear Equations in Two Unknowns
(a) Draw the graph of the equation 2y  x  2.
(b) Using the graph drawn in (a), answer the following questions.
(i) If P(–3, y) lies on the graph of 2y  x  2, find the value of y.
(ii) Is (3, 2.5) a solution of the equation 2y  x  2?
Solution:
2y  x  2
(b) (i) The x-coordinate of P is –3.
From the graph, the y-coordinate of P is –0.5.
(3, 2.5)
(ii) The point (3, 2.5) lies on the graph
of 2y  x + 2.
 (3, 2.5) is a solution of the equation 2y  x  2.
(3, 0.5)

15. 7.2 Solving Simultaneous Linear Equations in Two Unknowns by the Graphical Method
The solution obtained is an approximation only.

16. 7.2 Solving Simultaneous Linear Equations in Two Unknowns by the Graphical Method

17. Example 5T Solution: 7 Linear Equations in Two Unknowns
The figure shows the graph of the equation x  y  2.
(a) Complete the following table for the equation y  3x – 4.
(b) Draw the graph of y  3x – 4 in the same coordinate plane.
(c) Hence solve graphically. x 1 2
2.5 y
Solution:
(a) y  3x – 4
(b) x 1 2
2.5 y –1
3.5
y  3x  4
(c) From the graph, the approximate solution is x  1.5, y  0.5.
(1.5, 0.5)

18. Example 6T Solution: 7 Linear Equations in Two Unknowns
 x + 2y + 5  0
Solve the simultaneous equations  graphically.
4y – 3x  8
Give the answer correct to the nearest 0.1.
Solution: x –5 –3 –1 y –2 x –4 –2 y –1
0.5 2
From the graph, the approximate solution is x  –3.6, y  –0.7.

19. 7.3 Method of Substitution

20. 7.3 Method of Substitution

21. 7.3 Method of Substitution

22. 7.3 Method of Substitution

23. Example 7T Solution: 7 Linear Equations in Two Unknowns
Solve the simultaneous equations
Solution:
Substitute x  –1 into (1):

24. Example 8T Solution: 7 Linear Equations in Two Unknowns
Solve the simultaneous equations 2x + y + 1  x – y  5.
Solution:
Substitute y  –2 into (3):

25. Example 9T Solution: 7 Linear Equations in Two Unknowns
Solve the simultaneous equations
Solution:

26. 7.4 Method of Elimination
You can perform addition or subtraction in vertical form if you like.

27. Example 10T Solution: 7 Linear Equations in Two Unknowns
Solve the simultaneous equations by the method of elimination.
Solution:

28. 7.4 Method of Elimination

29. Example 11T Solution: 7 Linear Equations in Two Unknowns 3y  2x  47
Solve the simultaneous equations  by the method of elimination.
4y – x  48
Solution:
(2)  2: 8y – 2x  (3)
(1)  (3):
Substitute y  13 into (2):
(3y  2x)  (8y – 2x)  47  96
4(13) – x  48
3y  2x  8y – 2x  143
x  4
11y  143
∴ The solution is x  4, y  13.
y  13

30. Example 12T Solution: 7 Linear Equations in Two Unknowns 3x  2y  2
Solve the simultaneous equations  by the method of elimination.
4x – 5y  –28
Solution:
3x  2y  (1) 
4x – 5y  – (2)
(1)  4: 12x  8y  (3)
(2)  3: 12x – 15y  – (4)
(3) – (4):
Substitute y  4 into (1):
3x  2(4)  2
23y  92
x  –2
y  4
∴ The solution is x  –2, y  4.

31. 7.5 Applications of Simultaneous Linear Equations in Two Unknowns

32. 7.5 Applications of Simultaneous Linear Equations in Two Unknowns

33. Example 13T Solution: 7 Linear Equations in Two Unknowns
Lily has some $3 and $5 stamps. If the total number of stamps is 20 and their total value is $70, find the number of $3 and $5 stamps that she has.
Solution:
Let x and y be the numbers of $3 and $5 stamps respectively.
From the problem, we have
From (1), we have y  20 – x……(3)
Substitute (3) into (2): 3x  5(20 – x)  70
3x  100 – 5x  70
–2x  –30
x  15
Substitute x  15 into (3):
y  20 – 15  5
Lily has 15 $3 stamps and
5 $5 stamps.

34. Example 14T Solution: 7 Linear Equations in Two Unknowns
A company sells washing powder in 2 kinds of packages X and Y. The weight of one box of X is 50 g less than 2 boxes of Y. If the total weight of 2 boxes of X and 1 box of Y is 2 kg, find the weight of each package of washing powder.
Solution:
Let x g and y g be the weights of each package of X and Y respectively.
From the problem, we have
(1)  2: 4y – 2x  100……(3)
(2) + (3):
Substitute y  420 into (2):
The weights of each package of X and Y are 790 g and 420 g respectively.

35. Example 15T Solution: 7 Linear Equations in Two Unknowns
Dickson is 3 years older than his sister Mary. 2 years ago, Dickson was twice as old as Mary. How old is Dickson now?
Solution:
Let x and y be the present ages of Dickson and Mary respectively.
From the problem, we have
From (1), we have x  3  y……(3)
Substitute (3) into (2):
(3  y) – 2  2(y  2)
3  y – 2  2y – 4
–y  –5
y  5
Substitute y  5 into (3):
x  3  5  8
 Dickson is 8 years old now.

36. Example 16T Solution: 7 Linear Equations in Two Unknowns
Fanny finished a test with 2 parts A and B. She used of the time to
finish part A. If she used 20 minutes more to finish part A than part B,
(a) find the time she used to finish parts A and B respectively,
(b) find the time she used to finish the whole test.
Solution:
(a) Let x minutes and y minutes be the time she used to finish part A
and part B respectively.
From the problem, we have
From (2), we have
x  y + 20…….(3)
Substitute (3) into (1):

37. Example 16T Solution: 7 Linear Equations in Two Unknowns
Fanny finished a test with 2 parts A and B. She used of the time to
finish part A. If she used 20 minutes more to finish part A than part B,
(a) find the time she used to finish parts A and B respectively,
(b) find the time she used to finish the whole test.
Solution:
Substitute y  40 into (2):
x – 40  20
x  60
 The time she used to finish part A and part B are 60 minutes
and 40 minutes respectively.
(b) The time she used  (60  40) minutes

38. Follow-up 1 Solution: 7 Linear Equations in Two Unknowns
Determine whether the following ordered pairs are solutions of the equation x – 3y  1.
(a) (4, 1) (b) (1, 4)
Solution:
(a) Substitute (4, 1) into the equation.
(b) Substitute (1, 4) into the equation.
L.H.S.  4 – 3(1)
L.H.S.  1 – 3(4)
 4  3
 1  12
 1
 –11
 R.H.S.
 R.H.S.
∵ L.H.S.  R.H.S.
∵ L.H.S.  R.H.S.
 (4, 1) is a solution of the equation.
 (1, 4) is not a solution of the equation.

39. Follow-up 2 Solution: 7 Linear Equations in Two Unknowns
Does B(–4, 1) lie on the graphs of the following equations? Explain your answer.
(a) 2x – 3y = (b) 3x + 10y = –2
Solution:
(a) Substitute (–4, 1) into the equation 2x – 3y  4,
L.H.S.  2(–4) – 3(1)
 –11
 R.H.S.
(–4, 1) does not satisfy the equation 2x – 3y  4.
 B(–4, 1) does not lie on the graph of the equation 2x – 3y  4.

40. Follow-up 2 Solution: 7 Linear Equations in Two Unknowns
Does B(–4, 1) lie on the graphs of the following equations? Explain your answer.
(a) 2x – 3y = (b) 3x + 10y = –2
Solution:
(b) Substitute (–4, 1) into the equation 3x  10y  –2 ,
L.H.S.  3(–4)  10(1)
 –2
= R.H.S.
(–4, 1) satisfies the equation 3x  10y  –2.
 B(–4, 1) lies on the graph of the equation 3x  10y  –2.

41. Follow-up 3 Solution: 7 Linear Equations in Two Unknowns
The figure shows the graph of 9x  4y  36.
Find the coordinates of B.
Solution:
From the graph, the x-coordinate of B is 2.
Substitute x  2 into 9x  4y  36, we have
9(2)  4y  36
4y  18
y  4.5
∴ The coordinates of B are (2, 4.5).

42. Follow-up 4 Solution: 7 Linear Equations in Two Unknowns
Consider the equation y  1 – x.
(a) Draw the graph of y  1 – x on the given rectangular coordinate plane.
(b) Using the graph draw in (a), answer the following questions.
(i) If P(x, –0.5) lies on the graph of y  1 – x, find the value of x.
Solution:
(a) x –2 2 y 3 1 –1
(b) (i) The y-coordinate of P is 0.5.
From the graph, the x-coordinate of P is 1.5.
(1.5, –0.5)
 x  1.5
y  1 – x

43. Follow-up 4 Solution: 7 Linear Equations in Two Unknowns
Consider the equation y  1 – x.
(b) Using the graph draw in (a), answer the following questions.
(ii) If Q(–1.5, y) lies on the graph of y  1 – x, find the value of y.
(iii) Is R(2.5, –1) a solution of the equation y  1 – x? Explain your answer.
Solution:
(b) (ii) The x-coordinate of Q is 1.5.
(–1.5, 2.5)
From the graph, the y-coordinate of Q is 2.5.
 y  2.5
(iii)
The point R(2.5, 1) does not lie on the graph y  1  x.
(2.5, –1)
 (2.5, 1) is not a solution.
y  1 – x

44. Follow-up 5 Solution: 7 Linear Equations in Two Unknowns
The figure shows the graph of the equation 3x – 2y  –12.
(a) Complete the following table for the equation 2y – x  8.
(b) Draw the graph of 2y – x  8 in the same coordinate plane.
(c) Solve the simultaneous equations
graphically.
(–2, 3)
2y – x  8 x –5 –4 –1 y
Solution:
(a) x –5 –4 –1 y
1.5 2
3.5
(c) From the graph, the approximate solution is x  –2, y  3.

45. Follow-up 6 Solution: 7 Linear Equations in Two Unknowns
Solve the simultaneous equations graphically. Give the answer
correct to the nearest 0.1.
Solution:
y  2x  4 x 4 2 y –4
y  x – 1 x 4 2 y 3 1 –1
From to the graph, the approximate solution is x  1.7, y  0.7.

46. Follow-up 7 Solution: 7 Linear Equations in Two Unknowns
Solve the simultaneous equations
Solution:
From (2), we have y  2x  8……….(3)
Substitute (3) into (1):
Substitute x  3 into (3):
2x  3(2x  8)  12
y  2(3)  8
2x  6x  24  12
 2
4x  12
 The solution is x  3, y  2.
x  3

47. Follow-up 8 Solution: 7 Linear Equations in Two Unknowns
Solve the simultaneous equations y – 2x  2y – 7x  3.
Solution:
 y – 2x  (1) 
 2y – 7x  (2)
Rewriting the given equations, we have:
From (1), we have y  2x  (3)
Substitute (3) into (2):
2(2x  3) – 7x  3
Substitute x  1 into (3):
4x  6 – 7x  3
y  2(1)  3
 5
–3x  –3
 The solution is x  1, y  5.
x  1

48. Follow-up 9 Solution: 7 Linear Equations in Two Unknowns
Solve the simultaneous equations
Solution:
From (1), we have 3y  1 – 6x
y  (3)
Substitute x  –2 into (3):
Substitute (3) into (2):
5x  6　 　  16
5x  2 – 12x  16
–7x  14
x  –2
 The solution is x  –2, y  .

49. Follow-up 10 Solution: 7 Linear Equations in Two Unknowns
2x – 3y  5
Solve the simultaneous equations  by the method of elimination.
2x  y  9
Solution:
Substitute y  1 into (1):
2x – 3(1)  5
(2) – (1):
(2x  y) – (2x – 3y)  9 – 5
2x  8
2x  y – 2x  3y  4
x  4
4y  4
 The solution is x  4, y  1.
y  1

50. Follow-up 11 Solution: 7 Linear Equations in Two Unknowns
Solve the simultaneous equations by the method of elimination.
Solution:
(2)  2: 4x – 10y  –6…..(3)
(1)  (3):
Substitute y  1 into (1):

51. Follow-up 12 Solution: 7 Linear Equations in Two Unknowns
Solve the simultaneous equations by the method of elimination.
Solution:
(1)  3: 15x – 12y  6………(3)
(2)  4: 32x – 12y  108……(4)
(4) – (3):
Substitute x  6 into (1):

52. Follow-up 13 Solution: 7 Linear Equations in Two Unknowns
John has some $10 and $20 notes. If the total number of notes is 18 and their total value is $300, find the number of $10 and $20 notes that he has.
Solution:
Let x and y be the numbers of $10 and $20 notes respectively.
From the problem, we have
From (1), we have y  18 – x……(3)
Substitute (3) into (2):
Substitute x  6 into (3):
10x  20(18 – x)  300
10x  360 – 20x  300
–10x  –60
x  6
y  18 – 6
 12

53. Follow-up 14 Solution: 7 Linear Equations in Two Unknowns
Mrs. Lee bought some bottles of drinks X and Y. The total volume of 2 bottles of X and 1 bottles of Y is 4 L. If the total volume of 2 bottles of X is 1 L more than that of 2 bottles of Y, find the volume of each bottle of X and each bottle of Y respectively.
Solution:
Let x L and y L be the volumes of each bottle of X and each bottle of Y respectively. From the problem, we have
Substitute y  1 into (2):
(1) – (2):
The volumes of each bottle of X and each bottle of Y are 1.5 L and 1 L respectively.

54. Follow-up 15 Solution: 7 Linear Equations in Two Unknowns
Mrs. Wan is 30 years older than her daughter. 4 years ago, the total age of Mrs. Wan and her daughter was 34. How old is her daughter now?
Solution:
Let x and y be the present ages of Mrs. Wan and her daughter respectively.
From the problem, we have
Substitute (3) into (2):
[(30  y) – 4]  (y – 4)  34
30  y  4  y – 4  34
2y  12
y  6
From (1), we have
x  30  y……(3)

55. Follow-up 16 Solution: 7 Linear Equations in Two Unknowns
A box of sweets is shared among Jerry and Gloria in the ratio 5 : 2. If Jerry gets 60 more sweets than Gloria, find
(a) the number of sweets each of them gets,
(b) the total number of sweets in the box.
Solution:
(a) Let x and y be the numbers of sweets that Jerry and Gloria get
respectively. From the problem, we have
From (1), we have
Substitute (3) into (2):

56. Follow-up 16 Solution: 7 Linear Equations in Two Unknowns
A box of sweets is shared among Jerry and Gloria in the ratio 5 : 2. If Jerry gets 60 more sweets than Gloria, find
(a) the number of sweets each of them gets,
(b) the total number of sweets in the box.
Solution:
Substitute y  40 into (3):
(b) Total number of sweets in the box