Solve the equation x + 4 = 7 and check the solution.

##
x47
x4474
x3
Checking: Substitute x = 3 into both sides of the equation.
L.H.S. = 3 + 4 = 7
R.H.S. = 7
∵ L.H.S. = R.H.S.
∴ x = 3 is the solution.

========================================================
Solve the equation x – 6 = –14 and check the solution.

##
x  6  14
x  6  6  14  6
x  8
Checking: Substitute x = 8 into both sides of the equation.
L.H.S. = –8 – 6 = –14
R.H.S. = –14
∵ L.H.S. = R.H.S.
∴ x = 8 is the solution.

========================================================
Solve the equation 5y = 15 and check the solution.

##
5 y  15
5y 15

5 5
y3
Checking: Substitute y = 3 into both sides of the equation.
L.H.S. = 5(3) = 15
R.H.S. = 15
∵ L.H.S. = R.H.S.
∴ y = 3 is the solution.

========================================================
2y
Solve the equation  16 and check the solution.
5

##
 2y
 16
5
2y
 5  16  5
5
2 y  80
2y  80

2 2
y   40
Checking: Substitute y = 40 into both sides of the equation.
2( 40)
L.H.S. =  16
5
R.H.S. = 16
∵ L.H.S. = R.H.S.
∴ y = 40 is the solution.

========================================================
Solve the equation 3y + 60 = 48 and check the solution.

##
 3 y  60  48
 3 y  48  60
 3 y  12
 3 y  12

3 3
y4
Checking: Substitute y = 4 into both sides of the equation.
L.H.S. = 3(4) + 60 = 48
R.H.S. = 48
∵ L.H.S. = R.H.S.
∴ y = 4 is the solution.

========================================================
Solve the equation 4x – 7 = 33 and check the solution.

##
4 x  7  33
4 x  33  7
4 x  40
4 x 40

4 4
x  10
Checking: Substitute x = 10 into both sides of the equation.
L.H.S. = 4(10) – 7 = 33
 R.H.S. = 33
∵ L.H.S. = R.H.S.
∴ x = 10 is the solution.

========================================================
y
Solve the equation  4  10 and check the solution.
3

##
y
 4  10
3
y
 10  4
3
y
6
3
y
3  63
3
y  18
Checking: Substitute y = 18 into both sides of the equation.
18
L.H.S. =  4  10
3
R.H.S. = 10
∵ L.H.S. = R.H.S.
∴ y = 18 is the solution.

========================================================
s9
Solve the equation  18 and check the solution.
5

##
s9
 18
5
s9
 5  18  5
5
s  9  90
s  90  9
s   99
Checking: Substitute s = 99 into both sides of the equation.
 99  9
L.H.S. =  18
5
R.H.S. = 18
∵ L.H.S. = R.H.S.
∴ s = 99 is the solution.
========================================================
50  t
Solve the equation  6 and check the solution.
7

##
50  t
6
7
50  t
7  67
7
50  t  42
 t  42  50
 t  8
t 8
Checking: Substitute t = 8 into both sides of the equation.
50  8
L.H.S. = 6
7
R.H.S. = 6
∵ L.H.S. = R.H.S.
∴ t = 8 is the solution.

========================================================
Solve the equation 6 – (1 + x) = 2 and check the solution.

##
6  (1  x)  2
6 1 x  2
5 x  2
 x  25
 x  3
x3
Checking: Substitute x = 3 into both sides of the equation.
L.H.S. = 6 – (1 + 3) = 2
R.H.S. = 2
∵ L.H.S. = R.H.S.
∴ x = 3 is the solution.

========================================================
Solve the equation 7(5  z) = 14 and check the solution.

##
7(5  z )  14
35  7 z  14
 7 z  14  35
 7 z  21
 7 z  21

7 7
z 3
Checking: Substitute z = 3 into both sides of the equation.
L.H.S. = 7(5  3) = 14
R.H.S. = 14
∵ L.H.S. = R.H.S.
∴ z = 3 is the solution.

========================================================
2y 3
Solve the equation   3 and check the solution.
5 5

##
2y 3
 3
5 5
2y 3
 3
5 5
2 y 12

5 5
2 y  12
2 y 12

2 2
y6
Checking: Substitute y = 6 into both sides of the equation.
2(6) 3 12 3
L.H.S. =    3
5 5 5 5
R.H.S. = 3
∵ L.H.S. = R.H.S.
∴ y = 6 is the solution.

========================================================
m 1
Solve the equation   2 and check the solution.
3 6

##
m 1
 2
3 6
m 1
 2
3 6
m 13

3 6
m 13
3  3
3 6
13
m
2
13
Checking: Substitute m  into both sides of the equation.
2
 13 
 
L.H.S. =  2  1 13 1
   2
3 6 6 6
R.H.S. = 2
∵ L.H.S. = R.H.S.
13
∴ m is the solution.
2

========================================================
Solve the equation 4x + 2x = 36.

##
4x  2x  36
6x  36
6x 36

6 6
x 6

========================================================
Solve the equation 5y – 2y + 5 = 11.

##
5 y  2 y  5  11
3 y  5  11
3 y  11  5
3y  6
3y 6

3 3
y2
========================================================
Solve the equation 4 – 5z = 3z.

##
4  5 z  3z
 5 z  3 z  4
 8 z  4
 8z 4

8 8
1
z
2

========================================================
Solve the equation 15m + 2 = 22 – 5m.

##
15m  2  22  5m
15m  5m  22  2
20m  20
20m 20

20 20
m 1

========================================================
Solve the equation 3x + 24 = 11x – 8.

##
3 x  24  11 x  8
3 x  11 x  8  24
 8x  32
 8x  32

8 8
x 4

========================================================
Solve the equation 1.5k + 8.8 = 0.7k.

##
 1.5k  8.8  0.7 k
1.5k  0.7 k  8.8
2.2k  8.8
2.2k  8.8

2.2 2.2
k 4

========================================================
Solve the equation 2n + 2 = 1 – (4 – n).

##
2n  2  1  ( 4  n)
2n  2 1 4  n
2n  2  n3
2n  n  3  2
n  5

========================================================
Solve the equation x + 2(x + 1) = 4(x – 1).

##
x  2( x  1)  4( x  1)
x  2x  2  4x  4
3x  2  4x  4
3x  4 x  4  2
x  6
x  6

========================================================
4y 7y
Solve the equation   5.
3 3

##
4y 7y
 5
3 3
4y  7y
5
3
3y
 5
3
 y5
y  5

========================================================
 3t t
Solve the equation   1.
4 2

##
3t t
 1
4 2
 3t t 
    4  1 4
 4 2
3t t
4 4  4
4 2
3t  2t  4
t4

========================================================
5y 3y
Solve the equation 1  .
7 5

##
5y 3y
1 
7 5
 5y  3y
  1  35   35
 7  5
5y
 35  1  35  21 y
7
25 y  35  21 y
25 y  21 y  35
4 y  35
35
y
4

========================================================
Solve the equation 35  910p  310p .

##
 3 9p 3p
 
5 10 10
3 9p  3p
    10   10
 5 10  10
3 9p
 10   10  3 p
5 10
6  9 p  3p
 9 p  3 p  6
 12 p  6
 12 p 6

 12  12
1
p
2

========================================================
The sum of x times 5 and 2 is 4. Find the value of x.

##
According to the problem, we can set up an algebraic equation as follows:
5 x  2  4
5 x  4  2
5 x  6
6
x
5

========================================================
The product of 6 and a is greater than the sum of 6 and a by 39. Find the value of a.

##
According to the problem, we can set up an algebraic equation as follows:
6a  (6  a )  39
6a  a  39  6
5a  45
a9

========================================================
In the figure, the perimeter of the rectangle is 54. Find the value of w.

##
According to the problem, we can set up an algebraic equation as follows:
2( w  6)  54
2 w  12  54
2 w  54  12
2 w  42
w  21

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

In the figure, the perimeter of the quadrilateral is 25 cm. Find the value of n.

##
According to the problem, we can set up an algebraic equation as follows:
( 2n  2)  7  4  6  25
2n  15  25
2n  25  15
2n  10
n5

========================================================
Tom has $x. Jack has $50 more than twice the amount Tom has. If their sum is $170, how
much does Jack have?

##
Let $x be the amount Tom has. Then Jack has $(2x + 50).
x  (2 x  50)  170
∴ x  2 x  170  50
3x  120
x  40
∴ Tom has $40.
∴ The amount Jack has = $(2  40 + 50) = $130

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

Amy travels by a car of speed y km/h for 2 hours, then she walks for 5 km. If she travels
133 km in total, find the value of y.

##
According to the problem, we can set up an algebraic equation as follows:
2 y  5  133
2 y  133  5
2 y  128
y  64

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

Mabel weighs 71 kg and Carol weighs z kg. If Mabel’s weight is 40 kg less than 3 times
Carol’s weight, find the value of z.

##
According to the problem, we can set up an algebraic equation as follows:
3z  71  40
3z  40  71
3z  111
z  37

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

If each side of an equilateral triangle is decreased by 2 cm, the perimeter of the new
4
triangle is that of the original one. Find the length of a side of the original triangle.
5

##
Let x cm be the length of a side of the original triangle.
4
3( x  2)  (3 x )
5
4x
x2
5
5 x  10  4 x
5 x  4 x  10
x  10
∴ The length of a side of the original triangle is 10 cm.

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

A plant of 12.5 cm tall grows with a speed of 3.25 cm per week. How long does it take to
be 51.5 cm tall?

##
Let the plant takes x week to be 51.5 cm tall.
3.25 x  12.5  51.5
3.25 x  51.5  12.5
3.25 x  39
x  12
∴ The plant takes 12 weeks to be 51.5 cm tall.
========================================================

$140 is shared among A, B and C. If A gets $10 more than B and B gets $20 more than C,
find the amount each of them get.

##
Let $x be the amount A gets.
Then the amount B gets and C gets are $(x – 10) and $(x – 10 – 20), i.e. $(x – 30)
respectively.
x  ( x  10)  ( x  30)  140
x  x  x  140  10  30
3 x  180
x  60
∴ The amount A gets = $60
The amount B gets = $(60  10)  $50
The amount C gets = $(60  30)  $30

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

Solve the literal equation x + ab = c for x.

##
x + ab = c
∴ x  c  ab

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

3n
Solve the literal equation  c for n.
b

##
3n
c
b
3n  bc
∴ bc
n
3

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

Solve the literal equation 5y – a = 7b for y.

##
 5 y  a  7b
5 y  a  7b
∴ a  7b
y
5

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

Solve the literal equation 4z – vw = t for z.

##
4 z  vw  t
4 z  t  vw
∴ t  vw
z
4

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

m
Solve the literal equation  nx  q for x.
5

##
m
 nx  q
5
m
 nx  q 
5
∴ 1 m
x   q  
n 5 
q m
x 
n 5n