Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
1
Substitute (i) into (ii):
(y) + 3y =–4
4y =–4
−4
y
=
4
y
= –1
Substitute y =–1 into (i):
x
=–1
∴ The solution is x =–1, y =–1.
2
Adding the two equations:
(i)
3x–y = 2
(ii) 3x + y
=–2
(i) + (ii) 6x
=0
x
=0
Substitute x = 0 into (i):
3(0)–y
=2
–y
=2
y
=–2
∴ The solution is x = 0 , y =–2.
1
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
3
Substitute (ii) into (i):
(3b)–b
= 2.6
2b
= 2.6
b
=
2 .6
2
b
= 1.3
Substitute b = 1.3 into (ii):
a
= 3(1.3)
= 3.9
∴ The weights of the two goods are 3.9 kg and 1.3 kg respectively.
4
Substitute (i) into (ii):
1
y+4− y
2
=1
−
1
y+4=1
2
−
1
y
2
=–3
y
=6
Substitute y = 6 into (ii):
x–6 = 1
x
=7
∴ The solution is x = 7, y = 6.
2
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
5
(ii) 8x + 5y
=–1
(i)
=7
8x–3y
(ii)–(i)
8y =–8
y
=–1
Substitute y =–1 into (i):
8x–3(–1) = 7
8x + 3
=7
8x
=4
x
=
1
2
∴ The solution is x =
1
, y =–1.
2
6
(i) × 6
30x + 42y = 534 ............. (iii)
(ii) × 5
30x + 40y = 515 ............. (v)
(iii)–(v)
2y = 19
y = 9.5
Substitute y = 9.5 into (i):
5x + 7(9.5)
= 89
5x + 66.5 = 89
5x
= 22.5
x
= 4.5
∴ The price of one can of coke is $4.5 and the price of one bottle of orange
juice is $9.5.
3
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
7
From (i), 10–2y
=x+5
10–2y–5 = x
5–2y = x
i.e. x
= 5–2y .................... (iii)
Substitute (iii) into (ii):
2(5–2y) + y + 7
=5
10–4y + y + 7 = 5
–3y + 17 = 5
–3y =–12
y
=4
Substitute y = 4 into (iii):
x
= 5–2(4)\
=–3
∴ The solution is x =–3, y = 4.
4
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
8
The above equations can be written as
From (i) and (ii), we have
(iii) 9x–4y
(iv) × 2
=–1
6x + 4y
=–4 ....................... (v)
(iii) + (v) 15x =–5
x
= −
Substitute x = −
1
3( − ) + 2y
3
1
3
1
into (iv):
3
=–2
–1 + 2y =–2
2y
=–1
y
=−
1
2
1
1
∴ The solution is x = − , y = − .
3
2
5
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
9
(a)
From (i), (12)(
m n
+ )
3 4
= (12)(1)
4m + 3n = 12 ............... (iii)
From (ii), (6)(
m n
− )
2 3
= (6)(10)
3m–2n
= 60 ............... (iv)
(iii) × 2 8m + 6n = 24 .................... (v)
(iv) × 3
9m–6n
= 180 .................. (vi)
(v) + (vi) 17m = 204
m = 12
Substitute m = 12 into (iii):
4(12) + 3n
= 12
3n
=–36
n
=–12
∴ The solution is m = 12, n =–12.
(b)
From (i),
(2)(
x y
+
6 8
x y
+ )
6 8
x y
+
3 4
From (ii),
(20)(
=
= (2)(
1
2
1
)
2
=1
x
y
1
−
=
40 60
2
x
y
1
−
) = (20)( )
40 60
2
x y
−
2 3
= 10
By using the result in (a), the solution is x = 12, y =–12.
6
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
10
The above equations can be written as
From (i) and (ii), we have
(iii) 4p–2q
(iv) × 2
= 17
18p + 2q = 38 ....................... (v)
(iii) + (v) 22p = 55
p =
5
2
Substitute p =
5
into (iv):
2
5
9( ) + q = 19
2
45
+ q = 19
2
q
= −
7
2
∴ The solution is p =
11
5
7
, q =− .
2
2
Suppose Sam answers x questions correctly and y questions incorrectly.
(i) × 10
10x + 10y = 100 ............. (iii)
(ii) 20x–10y = 110
(iii) + (ii) 30x = 210
x =7
∴ Sam answers 7 questions correctly.
7
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
12
Let the original numbers of the black balls and the white balls be x and y
respectively. From the question, we have
From (ii),
x
y
( x − ) + ( y − ) = 57
6
2
5x y
+
6 2
(6)(
= 57
5x y
+ )
6 2
= (6)(57)
5x + 3y
= 342 ........ (iii)
3x–3y
= 42 ............... (iv)
(i) × 3
(iii) 5x + 3y
= 342
(iv) + (iii)
8x
= 384
x
= 48
Substitute x = 48 into (i):
48–y = 14
–y
=–34
y
= 34
∴ The original numbers of the black balls and the white balls are 48 and 34
respectively.
8
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
13.
KK (1)
2 x = 3 y

4 x + 5 y − 11 = 0 KK (2)
(2) – (1) × 2,
5 y − 11 = −6 y
y =1
By substituting y = 1 into (1), we have
2 x = 3(1)
x=
3
2
3
∴ The solution is x = , y = 1.
2
14.
3 x − 4 y + 1 = 0 KK (1)

5 x + y − 6 = 0 KK (2)
(1) + (2) × 4,
23 x − 23 = 0
x =1
By substituting x = 1 into (2), we have
5(1) + y − 6 = 0
y =1
∴ The solution is x = 1, y = 1.
15.
3( x + y ) = 2 y + 1

3(2 x + 3 y ) = −33
After simplification, we have
KK (1)
3 x + y − 1 = 0

2 x + 3 y + 11 = 0 KK (2)
(2) – (1) × 3,
− 7 x + 14 = 0
x=2
By substituting x = 2 into (1), we have
3(2) + y − 1 = 0
y = −5
∴ The solution is x = 2, y = –5.
9
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
16.
 2( x + y ) = y − 1

7(2 x + 3 y ) = −77
After simplification, we have
KK (1)
2 x + y + 1 = 0

2 x + 3 y + 11 = 0 KK(2)
(2) − (1),
2 y + 10 = 0
y = −5
By substituting y = −5 into (1), we have
2 x + (−5) + 1 = 0
x=2
∴ The solution is x = 2, y = –5.
17.
 x + y = 10

 3x y
 2 − 4 = 1
After simplification, we have
 x + y − 10 = 0 KK (1)

6 x − y − 4 = 0 KK (2)
(1) + (2),
7 x − 14 = 0
x=2
By substituting x = 2 into (1), we have
2 + y − 10 = 0
y =8
∴ The solution is x = 2, y = 8.
18.
0.5 x + 0.25 y = 1.5

1.5 x − 2.5 y = 4.5
KK (1)
K K ( 2)
(2) – (1) × 3,
− 3.25 y = 0
y=0
By substituting y = 0 into (1), we have
0.5 x + 0 = 1.5
x=3
∴ The solution is x = 3, y = 0.
10
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
19.
KK (1)
25 y + 8 x = 40

0.5 y + 0.4 x = 1.4 KK (2)
(2) × 20,
10y + 8x = 28 KK (3)
(1) – (3),
15 y = 12
y=
4
5
By substituting y =
4
into (3), we have
5
4
10  + 8 x = 28
5
8 x = 20
5
2
5
4
The solution is x = , y = .
2
5
x=
∴
20.
KK (1)
6 x + 5 y = 1

4
8

2 x − 3 y = − 3 KK (2)

(1) – (2) × 3,
9y = 9
y =1
By substituting y = 1 into (1), we have
6 x + 5(1) = 1
6 x = −4
2
x=−
3
∴ The solution is x = −
2
, y = 1.
3
11
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
21.
x−4

6 y − 3 = 10

x + 3y − 7 = 1
4

After simplification, we have
− x + 18 y − 26 = 0 KK (1)

K K ( 2)
4 x + 3 y − 11 = 0
(1) – (2) × 6,
− 25 x + 40 = 0
x=
8
5
By substituting x =
8
into (1), we have
5
8
+ 18 y − 26 = 0
5
23
y=
15
8
23
The solution is x = , y = .
5
15
−
∴
22.
y−6

4
x
−
= 18

3

 1 y + 4 x − 12 = 8
 3
2
After simplification, we have
 12 x − y − 48 = 0 KK (1)

2 y + 12 x − 84 = 0 KK (2)
(2) − (1),
3 y − 36 = 0
3 y = 36
y = 12
By substituting y = 12 into (1), we have
12 x − 12 − 48 = 0
12 x = 60
∴
x=5
The solution is x = 5, y = 12.
12
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
23.
2(b – a) + (3a – 2b) = 2b – 2a + 3a – 2b
=a
a
=
4
−
3
a
......(
1
)
∴ 

a = b − 2 ......(2)
From (1), we have
4a = 4
a =1
By substituting a = 1 into (2), we have
1= b − 2
b=3
∴ The solution is a = 1, b = 3.
∵
24.
(a)
4 x + 3 y = 17 KK (1)

K K ( 2)
 x − y = −1
(1) + (2) × 3,
7 x = 14
x=2
By substituting x = 2 into (2), we have
2 − y = −1
y=3
∴ The solution is x = 2, y = 3.
1
1
and b = .
(b) Let a =
x
y
The given equations can be written as
4a + 3b = 17

 a − b = −1
By (a), a = 2 and b = 3
1
1
∴
= 2, = 3
x
y
1
1
∴ x= , y=
2
3
13
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
25. (a)
x + 2 y = 2

3 x − 4 y = 1
KK (1)
KK ( 2)
(1) × 2 + (2),
5x = 5
x =1
By substituting x = 1 into (1), we have
1 + 2y = 2
y=
1
2
∴ The solution is x = 1, y =
1
.
2
1
1
and n = .
y
x
1
1
∴ x=
and y =
n
m
The given equations can be written as
x + 2 y = 2

3 x − 4 y = 1
1
By (a), x = 1, y =
2
1
1
∴ m = = 1, n =
=2
1
1
 
2
(b) Let m =
26. (a)
......(1)
x + 2 y = 3

6 x − 13 y = −32 ......(2)
From (1), we have x = 3 – 2y ……(3)
6(3 − 2 y ) − 13 y = −32
18 − 12 y − 13 y = −32
− 25 y = −50
y=2
x = 3 − 2(2)
∴
= −1
The solution is x = –1, y = 2.
(b) Let m = x – 1 and n = 4 – y.
∴ x = m + 1, y = 4 – n
From the result obtained in (a), we have
m + 1 = −1 and 4 − n = 2
m = −2 and
n=2
∴ The solution is m = –2, n = 2.
14
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
27. ∵
x=y
∴ The equations become
4 x + 3 x = 1

kx + (k − 1) x = 3
KK (1)
7 x = 1

(2k − 1) x = 3 KK (2)
1
By (1), x =
7
1
into (2), we have
By substituting x =
7
(2k − 1)
=3
7
2k = 22
k = 11
28. By substituting (2, k) into 8x + 2y + b = 0 and x – y + 2b – 1 = 0, we have
16 + 2k + b = 0 ......(1)

2 − k + 2b − 1 = 0 ......(2)
From (2), we have k = 2b + 1 ……(3)
By substituting (3) into (1), we have
16 + 2(2b + 1) + b = 0
16 + 4b + 2 + b = 0
5b = −18
18
b=−
5
By substituting b = −
18
into (3), we have
5
 18 
k = 2 −  + 1
 5
31
=−
5
15
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
29. Let x and y be the tens digit and the units digit of the number respectively.
 x + y = 11

10 y + x = 10 x + y − 45
After simplification, we have
 x + y = 11 KK (1)

 x − y = 5 K K ( 2)
(1) + (2),
2 x = 16
x=8
By substituting x = 8 into (1), we have
8 + y = 11
y=3
∴ The original number is 83.
30. Let the original fraction be
x
.
y
From the question, we have
x +1 3
y +1 = 4


x − 5 = 1
 y − 5 2
After simplification, we have
4( x + 1) = 3( y + 1)

2( x − 5) = y − 5
4 x − 3 y + 1 = 0 KK (1)
∴ 
 2 x − y − 5 = 0 K K ( 2)
(1) – (2) × 2,
− y + 11 = 0
y = 11
By substituting y = 11 into (2), we have
2 x − 11 − 5 = 0
x=8
8
∴ The original fraction is
.
11
16
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
31. Let x and y be the hundreds digit and the units digit of the number respectively.
From the question, we have
 x + y = 11

100 y + 70 + x = 100 x + 70 + y + 693
After simplification, we have
 x + y = 11 KK (1)

 y − x = 7 KK (2)
(1) + (2),
2 y = 18
y=9
By substituting y = 9 into (1), we have
x + 9 = 11
x=2
∴ The original number is 279.
32. Let x kg of alloy A and y kg of alloy B are needed.
0.25 x + 0.375 y = 0.3 × 100
From the question, we have 
0.75 x + 0.625 y = 0.7 × 100
After simplification, we have
2 x + 3 y = 240 KK (1)

6 x + 5 y = 560 KK (2)
(2) – (1) × 3,
− 4 y = −160
y = 40
By substituting y = 40 into (1), we have
2 x + 3(40) = 240
x = 60
∴ The craftsman needs 60 kg of alloy A and 40 kg of alloy B.
17
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
33. Suppose x mL of drink A and y mL of drink B are needed.
0.15 x + 0.05 y = 0.08 × 250
From the question, we have 
0.85 x + 0.95 y = 0.92 × 250
After simplification, we have
3 x + y = 400 KK (1)

17 x + 19 y = 4600 KK (2)
(1) × 19, 57x + 19y = 7600 KK (3)
(3) – (2),
40 x = 3000
x = 75
By substituting x = 75 into (1), we have
3(75) + y = 400
y = 175
∴ Steve needs 75 mL of drink A and 175 mL of drink B.
10m − 10n = 20 KK (1)
34. From the question, we have 
15m − 5n = 60 KK (2)
(1)  (2) × 2,
− 20m = −100
m=5
By substituting m = 5 into (1), we have
10(5) − 10n = 20
n=3
35. Let x km/h and y km/h be the speeds of cars A and B respectively.
From the question, we have
8
 60 ( x + y ) = 10

 40 ( x − y ) = 10
 60
After simplification, we have
 x + y = 75 KK (1)

 x − y = 15 KK (2)
(1) + (2),
2 x = 90
x = 45
By substituting x = 45 into (1), we have
18
Book 2B: Chapter 8 – Linear Equations in Two Unknowns (solution)
45 + y = 75
y = 30
∴ The speeds of car A and car B are 45 km/h and 30 km/h respectively.
40 x + 60 y = 1800
36. 
50( x + y − 2) = 1800
After simplification, we have
2 x + 3 y = 90 LL (1)

LL (2)
 x + y = 38
(2) × 2,
2 x + 2 y = 76 LL (3)
(1) – (3),
(2x + 3y) − (2x + 2y) = 90 − 76
y = 14
By substituting y = 14 into (2), we have
x + 14 = 38
x = 24
19