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Chapter 9: Introduction to Algebra > Exercise 9.3

Exercise 9.3

1. State which of the following are equations.

x – 3 = 7
l + 5 > 9
p – 4 < 10
5 + m = – 6
2s – 2 = 12
3x + 5 > 13
3x < 15
2x – 5 = 3
7y + 1 < 22
–3z + 6 = 12
2x – 3y = 3
z = 4
Equation
Not an Equation

2. Write LHS and RHS of the following equations.

Solution :

EquationLHS (Left-Hand Side)RHS (Right-Hand Side)
(i) x – 5 = 6x – 5
(ii) 4y = 124y
(iii) 2z + 3 = 7 + 7
(iv) 3p = 24
(v) 4 = x – 24
(vi) 2a – 3 = –5

3(i) .Solve the following equations by Trial & Error method.

(i) x + 3 = 5

Solution :

Value of xValue of LHSValue of RHSWhether LHS and RHS are equal
1 + 3 =
2 + 3 =

We find that for x = , both LHS and RHS are equal. Therefore, x = 2 is the solution of the equation.

3(ii) Solve the following equations by Trial & Error method.

(ii) y – 2 = 7

Value of yValue of LHSValue of RHSWhether LHS and RHS are equal
11 – 2 =
22 – 2 =
33 – 2 =
44 – 2 =
55 – 2 =
66 – 2 =
77 – 2 =
88 – 2 =
99 – 2 =

We find that for y = , both LHS and RHS are equal. Therefore, y = 9 is the solution of the equation.

3(iii) Solve the following equations by Trial & Error method.

(iii) a – 2 = 6

Value of aValue of LHSValue of RHSWhether LHS and RHS are equal
11 – 2 =
22 – 2 =
33 – 2 =
44 – 2 =
55 – 2 =
66 – 2 =
77 – 2 =
88 – 2 =

We find that for a = , both LHS and RHS are equal. Therefore, a = 8 is the solution of the equation.

Solve the following equations by Trial & Error method.

(iv) 5y = 15

Value of yValue of LHSValue of RHSWhether LHS and RHS are equal
15 × 1 =
25 × 2 =
35 × 3 =

We find that for y = , both LHS and RHS are equal. Therefore, y = 3 is the solution of the equation.

(v) 6n = 30

Value of nValue of LHSValue of RHSWhether LHS and RHS are equal
16 × 1 =
26 × 2 =
36 × 3 =
46 × 4 =
56 × 5 =

We find that for n = , both LHS and RHS are equal. Therefore, n = 5 is the solution of the equation.

(vi) 3z = 27

Value of zValue of LHSValue of RHSWhether LHS and RHS are equal
13 × 1 =
23 × 2 =
33 × 3 =
43 × 4 =
53 × 5 =
63 × 6 =
73 × 7 =
83 × 8 =
93 × 9 =

We find that for z = , both LHS and RHS are equal. Therefore, z = 9 is the solution of the equation.