Solve the following equations and check the solutions.

(a) x  7  15
1
(b) x  9  14
2

##
x  7  15
(a) x  7  7  15  7
x  8
Checking: Substitute x = –8 into both sides of the equation.
L.H.S. = –8 – 7 = –15
R.H.S. = –15
∴ x = –8 is the solution.

1
x  9  14
2
(b) 1
x  9  9  14  9
2
1
x5
2
1
Checking: Substitute x  5 into both sides of the equation.
2
1 1
L.H.S.  5  9  14
2 2
1
R.H.S.  14
2
1
∴ x  5 is the solution.
2
======================================================

(k  5) x (m  2) x
If x is an unknown, solve the literal equation   5.
3 4

##
(k  5) x (m  2) x
 5
3 4
 (k  5) x (m  2) x 
 3   12  5  12
4 
4(k  5) x  3(m  2) x  60
(4k  20) x  (3m  6) x  60
[(4k  20)  (3m  6)]x  60
(4k  3m  26) x  60
60
x
4k  3m  26

1
======================================================
Solve the following equations and check the solutions.
(a)  4a  28
b
(b)  5
6

##
 4a  28
(a)  4a 28

4 4
a  7
Checking: Substitute a = –7 into both sides of the equation.
L.H.S. = –4(–7) = 28
R.H.S. = 28
∴ a = –7 is the solution.

b
 5
6
(b) b
 6  5  6
6
b   30
Checking: Substitute b = –30 into both sides of the equation.
 30
L.H.S.   5
6
R.H.S. = –5
∴ b = –30 is the solution.

======================================================

Solve the following equations and check the solutions.

(a) 5a  7  22
7  2y
(b) 1
5

##
5a  7  22
5a  22  7
(a) 5a  15
5a 15

5 5
a3
Checking: Substitute a = 3 into both sides of the equation.
L.H.S. = 5(3) + 7 = 22
R.H.S. = 22
∴ a = 3 is the solution.

2
 7  2y
1
5
7  2y
 5  1 5
5
(b) 7  2y  5
 2y  5  7
 2 y  2
 2y 2

2 2
y 1
Checking: Substitute y = 1 into both sides of the equation.
7  2(1)
L.H.S.  1
5
R.H.S. = 1
∴ y = 1 is the solution.

======================================================

Solve the following equations.

(a) 7 y  4 y  30
(b) 5a  18  3a  10

##
7 y  4 y  30
3 y  30
(a) 30
y
3
y  10

5a  18  3a  10
2a  18  10
2a  10  18
(b) 2a  28
 28
a
2
a   14

======================================================

Solve the equation 6n  8  56  2n .

##

3
 6n  8  56  2n
6n  2n  8  56
8n  8  56
8n  56  8
8n  48
48
n
8
n6

======================================================

Solve the following equations.

(a) 3 p  2  5(8  p )
(b) 7 ( 7  q )  4( q  4)

##
3 p  2  5(8  p)
3 p  2  40  5 p
3 p  5 p  2  40
 2 p  2  40
(a)  2 p  40  2
 2 p  42
42
p
2
p   21

7(7  q)  4(q  4)
49  7q  4q  16
49  7 q  4q  16
49  11q  16
(b)  11q  16  49
 11q  33
 33
q
 11
q3

======================================================

Solve the equation 6( y  4)  5[17  3(2 y  13)] .

##

4
 6( y  4)  5[17  3( 2 y  13)]
6 y  24  5(17  6 y  39)
6 y  24  5(56  6 y )
6 y  24  280  30 y
6 y  30 y  24  280
36 y  24  280
36 y  280  24
36 y  304
304
y
36
4
y 8
9

======================================================

Solve the following equations.

x 2x 4
(a)  
3 5 15
3y  4 5 y  7
(b)  
4 6 3

##
x 2x 4
 
3 5 15
 x 2x  4
    15   15
 3 5  15
(a) x 2x
 15   15  4
3 5
5x  6 x  4
x4
x  4

3y  4 5 y  7
 
4 6 3
 3y  4 5  y7
    12   12
 4 6  3
3y  4 5
(b)  12   12  4( y  7)
4 6
3(3 y  4)  10  4( y  7)
9 y  12  10  4 y  28
9 y  4 y  28  12  10
5 y  30
y6

======================================================

5
 5(3 x  1)  1
Solve the equation  11  2 x   .
2  3

##
5(3 x  1)  1
 11  2 x  
2  3
15 x  5 2
 11  2 x 
2 3
 15 x  5   2
  11   6   2 x    6
 2   3
3(15 x  5)  66  12 x  4
45 x  15  66  12 x  4
45 x  12 x  4  51
33 x  55
55
x
33
5
x
3

======================================================

Mary has $x. She spends $80 to buy a dress, and has $50 left. Find the value of x.

##
According to the problem, we can set up an algebraic equation as follows:
x  80  50
x  50  80
x  130

======================================================

There are some red balls and green balls. If the number of red balls is greater than the
number of green balls by 22, and the total number of balls is 50, find the number of red
balls.

##
Let x be the number of red balls. Then the number of green balls is x – 22.
x  ( x  22)  50
2 x  50  22
∴ 2 x  72
72
x
2
x  36
∴ The number of red balls is 36.

6
======================================================

Tom buys a monitor and a scanner for $2100. If the price of the monitor is $300 more
than twice that of the scanner, find the price of the scanner.

##
Let $x be the price of the scanner. Then the price of the monitor is $(2x + 300).
(2 x  300)  x  2100
∴ 3 x  2100  300
3 x  1800
x  600
∴ The price of the scanner is $600.

======================================================

There are some 50 g weights and 100 g weights, and their total weight is 1550 g. If the
number of 100 g weights is 2 less than three times that of 50 g weights, find the number
of 50 g weights.

##
Let x be the number of 50 g weights. Then the number of 100 g weights is 3x – 2.
50 x  100(3 x  2)  1550
50 x  300 x  200  1550
350 x  1550  200
∴
350 x  1750
1750
x
350
x5
∴ The number of 50 g weights is 5.

======================================================

7c  ax
If x is an unknown, solve the literal equation d.
e

##
7c  ax
d
e
7c  ax  de
∴ ax  de  7c
de  7c
x
a

======================================================

7
 1 y b
Solve the literal equation for y:     y.
2 a 3

##
1 y b
   y
2 a 3
y b
 y
2a 6
y b
y
2a 6
1  b
  1 y 
 2a  6
∴
 1  2a  b
 y 
 2a  6
2ab
y
6(1  2a)
ab
y
3(1  2a)
ab
y
3  6a

======================================================

Solve the following equations and check the solutions.

 3x 
(a) 5 2    55
 4 
 14  3 y  1
(b) 4  1
 15  3

##

8
  3x 
5 2    55
 4
 3x 
5 2  
 4  55

5 5
3x
2  11
(a) 4
3x
 11  2
4
3x
9
4
3x 4 4
  9
4 3 3
x  12
Checking: Substitute x = 12 into both sides of the equation.
 3(12) 
L.H.S.  52   5(2  9)  55
 4 
R.H.S. = 55
∴ x = 12 is the solution.

 14  3 y  1
4  1
 15  3
56  12 y 4

15 3
56  12 y 4
 15   15
(b) 15 3
56  12 y  20
 12 y  20  56
 12 y  36
 36
y
 12
y3
Checking: Substitute y = 3 into both sides of the equation.
14  3(3)   5 1
L.H.S.  4    4   1
 15   15  3
1
R.H.S.  1
3
∴ y = 3 is the solution.

======================================================

Solve the equation 2 y  (10  3 y )  25 .

##

9
2 y  (10  3 y )  25
2 y  10  3 y  25
5 y  10  25
5 y  25  10
5 y  35
35
y
5
y7

======================================================

5  6( r  2)
Solve the equation 3  2r  .
3

##
5  6(r  2)
3  2r 
3
3(3  2r )  5  6( r  2)
9  6r  5  6r  12
9  6r  6r  7
6r  6 r  7  9
12r  16
 16
r 
12
4
r 
3

======================================================

12  [4  2( y  1)]
Solve the equation  2y  6 .
2

##
12  [4  2( y  1)]
 2y  6
2
12  ( 4  2 y  2)  2(2 y  6)
12  ( 2  2 y )  4 y  12
12  2  2 y  4 y  12
10  2 y  4 y  12
2 y  4 y  12  10
 2y  2
2
y
2
y  1

10
======================================================

t  1 3(t  1) 5(t  1)
Solve the equation   .
6 8 12

##
t  1 3(t  1) 5(t  1)
 
6 8 12
t  1 3(t  1)  5(t  1)
 6   8   24  12  24
4(t  1)  9(t  1)  10(t  1)
4t  4  9t  9  10t  10
13t  5  10t  10
13t  10t  10  5
3t  15
t 5

======================================================

A father is 27 years older than his son. Five years later, the sum of their ages will be 49.
Find their present ages.

##
Let x be the son’s present age. Then the father’s present age is x + 27.
( x  5)  [( x  27)  5]  49
x  5  x  27  5  49
∴ 2 x  37  49
2 x  49  37
2 x  12
x6
∴ The son’s present age is 6 years old and the father’s present age is 6 + 27 = 33
years old.

======================================================

3 5
Jack bought a cake and ate of it. May ate of the remaining and 450 g of the
10 14
cake was left. Find the original weight of the cake.

##
3x 5  3x 
Let x g be the original weight of the cake. Jack ate g and May ate  x   g.
10 14  10 

11
 3 5  3 
x x  x  x   450
10 14  10 
3x 5x 3x
x    450
10 14 28
 3 5 3 
1     x  450
 10 14 28 
9
x  450
20
450
x
9
20
x  1000
∴ The original weight of the cake is 1000 g.

======================================================

David has 34 stamps and there are three types of stamps: $1.4 stamps, $2.5 stamps and
$3.1 stamps. The number of $1.4 stamps is 2 more than three times that of $3.1 stamps.
The number of $2.5 stamps is more than the number of $3.1 stamps by 7. Find the total
value of the stamps.

##
Let x be the number of $3.1 stamps. Then the number of $1.4 stamps is 3x + 2 and the
number of $2.5 stamps is x + 7.
x  (3 x  2)  ( x  7)  34
5 x  9  34
∴ 5 x  34  9
5 x  25
x5
∴ There are 5 $3.1 stamps, 17 $1.4 stamps and 12 $2.5 stamps.
∴ The total value of the stamps
 $(5  3.1  17  1.4  12  2.5)
 $(15.5  23.8  30)
 $69.3

12