Students can Download Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 7 Maths Chapter 4 Simple Equations Ex 4.2

Question 1.

Given first the step you will use to separate the variable and then solve the equation:

a) x – 1 = 0

Solution:

The given equation = x – 1 = 0

Add 1 to both the sides

x – 1 + 1 = 0 + 1 ⇒ x = 1

Checking

x – 1 = 0

put the value of x = 1

1 – 1 = 0

∴ LHS = RHS (checked)

b) x + 1 = 0

The given equation = x + 1 = 0

Subtract 1 from both the sides

x – 1 + 1 = 0 – 1 ⇒ x = – 1

Verification

x+ 1 = 0

substitute the value of x = – 1 then x + 1 = 0

(-1) + 1 = 0

0 = 0

LHS = RHS (verified)

c) x – 1 = 5

The given equation is x – 1 = 5

Add 1 to both sides

x – 1 + 1 = 5 + 1 ⇒ x = 6

Substitute the value in the equation x = 6.

x – 1 = 5

6 – 1 = 5

5 = 5

∴ LHS = RHS (verified)

d) x + 6 = 2

The given equation is x + 6 = 2

Subtract 6 from both the sides

x + 6 – 6 = 2 – 6 ⇒ x = – 4

Substitute the value of x = – 4 in the given equation,

x + 6 = 2

– 4 + 6 = 2

2 = 2

∴ LHS = RHS (verified)

e) y – 4 = -7

The given equation is y – 4 = – 7

Add 4 to both the sides

y – 4 + 4 = – 7 + 4 ⇒ y = – 3

substitute the value of y = – 3 in the given equation

y – 4 = -7

– 3 – 4 = – 7

– 7 = – 7

∴ LHS = RHS (verified)

f) y – 4 = 4

Solution:

The given equation is y – 4 = 4

Add 4 to both the sides

y – 4 + 4 = 4 + 4 ⇒ y = 8

substitute the value of y = 8 in the given equation

y – 4 = 4

8 – 4 = 4

4 = 4

∴ LHS = RHS (verified)

g) y + 4 = 4

Solution:

The given equation is y + 4 = 4

subtract 4 from both sides

= y – 4 + 4 = 4 – 4 ⇒ y = 0

substitute the value of y = 0 in the given equation.

y + 4 = 4

0 + 4 = 4

4 = 4

∴ LHS = RHS (verified)

h) y + 4 = – 4

Solution:

The given equation is y + 4 = – 4

subtract 4 from both the sides

y – 4 + 4 = – 4 – 4

y = – 8 (verified)

substitute the value of y = – 8 in the given equation

y + 4 = – 4

– 8 + 4 = – 4

– 4 = – 4

∴ LHS = RHS (verified)

Question 2.

Give first the step you will use to separate the variable and then solve the equation:

a) 3l = 42

Solution:

The given equation is 3l = 42

Divide by 3 both the sides

\(\frac{31}{3}\) = \(\frac{42}{3}\)

∴ l = 14

Substitute the value of l = 14 in the given equation

3l = 42

3(14) = 42

42 = 42

∴ LHS = RHS (verified)

b) \(\frac{\mathbf{b}}{2}\) = 6

Solution:

The given equation is \(\frac{\mathbf{b}}{2}\) = 6

Multiplied by 2 both the sides b

\(\frac{b}{2}\) × 2 = 6 × 2

∴ b = 12

substitute the value of b = 12 in the given equation

∴ LHS = RHS (verified)

c) \(\frac{\mathbf{p}}{7}\) = 4

Solution:

The given equation is \(\frac{p}{7}\) = 4

Multiplied by 7 to both the sides

\(\frac{p}{7}\) ×7 = 4 × 7

∴ p = 28

Substitute the value of p = 28 in the given equation

∴ LHS = RHS (verified)

d) 4x = 25

Solution:

The given equation is 4 x = 25

Divide both sides by 4

substitue the value of x = \(\frac{25}{4}\) in the given equation

4x = 25

4 × \(\frac{25}{4}\) = 25

∴ LHS = RHS (verified)

e) 8y = 36

Solution:

The given equation is 8y = 36

Divide both sides by 8

Substitute the equation by y = \(\frac{9}{2}\) in the given equation

8y = 36

∴ LHS = RHS

f) \(\frac{z}{3}=\frac{5}{4}\)

Solution:

Multiply both sides by 3

g) \(\frac{a}{5}=\frac{7}{15}\)

Solution:

Multiply both sides by 5

Substitute the value of a = \(\frac{7}{3}\) in the given equation

h) 20t = – 10

Solution:

The given equation is 20 t = – 10

Divide by 20 to both sides

Question 3.

Given the steps you will use to separate the variable and then solve the equation:

a) 3n – 2 = 46

Solution:

Add 2 to both sides 3n – 2 + 2 = 46 + 2

3n = 48

Divide the equation by 3

\(\frac{3n}{3}\) = \(\frac{48}{3}\)

∴ n = 16 This is required solution.

substitute the value of n = 16 in the given equation

3n – 2 = 46

3(16) – 2 = 46

48 – 2 = 46

46 = 46

∴ LHS = RHS (verified)

b) 5m + 7 = 17

The given equation is 5m + 7 = 17

Subtract 7 from both sides

5m + 7 – 7 = 17 – 7

5m = 10

Divide the equation by 5 both the sides

\(\frac{5m}{5}\) = \(\frac{10}{5}\)

∴ m = 2 (solution)

Substitute the value of m = 2 in the given equation

5m + 7 = 17

5(2) + 7 =17

10 + 7 = 17

∴ LHS = RHS (verified)

c) \(\frac{20 p}{3}\) = 40

Solution:

Multiply both sides by 3

\(\frac{20}{3}\) × 3 × P = 40 × 3

20P = 120

Divide both sides by 20

\(\frac{20}{20}\) × P = \(\frac{120}{20}\) ⇒ P = 6

d) \(\frac{3 P}{10}=6\)

Solution:

Multiply by 10 to both the sides

Divide by 3 to both sides

\(\frac{3}{3}\) × P = \(\frac{60}{3}\) ⇒ P = 20

Question 4.

Solve the following equations :

a) 10p = 100

Solution:

b) 10p + 10 = 100

Solution:

10 P + 10 – 10 = 100 – 10

10 P = 90

c) \(\frac{\mathbf{p}}{4}\) = 5

Solution:

\(\frac{p}{4}\) × 4 = 5 × 4 ⇒ P = 20

d) \(\frac{-p}{3}\) = 5

Solution:

– P = 15 ⇒ P = – 15

e) \(\frac{3 P}{4}=6\)

Solution:

f) 3s = – 9

Solution:

\(\frac{3s}{3}=\frac{-9}{3}\)

S = – 3

g) 3s + 12 = 0

Solution:

3S + 12 + 12 – 12 = 0 – 12

h) 3s = 0

Solution:

j) 2q – 6 = 0

Solution:

2q – 6 + 6 = 0 + 6

k) 2q + 6 = 0

Solution:

2q + 6 + 6 = 0 – 6

2q = – 6

l) 2q + 6 = 12

2q + 6 – 6 = 12 – 6

2q = 6