Students can Download Chapter 4 Simple Equations Ex 4.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## 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

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

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

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

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

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

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

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 = -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

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

substitute the value of b = 12 in the given equation

∴ LHS = RHS (verified)

c)

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

Solution:

Multiplied by 7 to both the sides

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

∴ 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)

Solution:

Substitute the equation by z = \(\frac{15}{4}\) in the given equation

∴ LHS = RHS (verified)

g)

Solution:

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

∴ LHS = RHS (verified)

h) 20t = -10

Solution:

The given equation is 20 t = -10

Divide by 20 to both sides

LHS = RHS (verified)

Question 3.

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

a) 3n – 2 = 46

Solution:

The given equation is 3n – 2 = 46

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

3n = 48

Divide the equation by 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 = 10

Divide the equation by 5 both the sides

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:

Substitute the value of p = 6 in the given equation. .

40 = 40

∴ LHS = RHS (verified)

d)

Solution:

Multiply by 10 to both the sides

Divide by 3 to both sides

Substitute the value of p = 20 in the given equation

Question 4.

Solve the following equations :

a) 10p = 100

Solution:

The given equation is 10p = 100

Divide by 10 both the sides

Substitute the value of p in the given equation

10p = 100

10 x 10 = 100

100 = 100

∴ LHS = RHS (verified)

b) 10p + 10 = 100

Solution:

The given equation is 10p + 10 = 100

subtract 10 from both the sides

Divide by 10 both the sides

Substitute the value of p = 9 in the given equation

10p + 10 = 100

10 × 9 + 10 = 100

90 + 10 = 100

100 = 100

LHS = RHS (verified)

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

Solution:

5 = 5

∴ LHS = RHS (verified)

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

The given equation is \(\frac{-p}{3}\) = 5

The equation is multiplied by -3 both the sides

Substitute the value of p = -15 in the given equation

∴ LHS = RHS (verified)

e)

Solution:

Substitute the value p = 8 in the given equation

f) 3s = -9

The given equation is 3s = -9

Divide the equation by 3 both the sides

Substitute the value of s = -3 in the given equation

3s = -9

3(-3)=-9

-9 = -9

∴ LHS = RHS (verified)

g) 3s + 12 = 0

Solution:

The given equation is 3s + 12 = 0

subtract 12 from both the sides

Divide by 3 both the sides

Substitute the value of s = -4 in the given equation

3s = 12 = 0

3(-4) = 12 = 0

-12 + 12 = 0

0 = 0

∴ LHS = RHS

h) 3s = 0

Solution:

The given equation is 3s = 0

Divide by 3 both the sides

Substitute the value of s = 0 in the given equation

3s = 0

3(0) = 0

0 = 0

∴ LHS = RHS (verified)

i) 2q = 6

Solution:

The given equation is 2q = 6

Divide the equation by 2 both the sides

substitute the value of q = 3 in the given equation

2q = 6

2(3) = 6 6 = 6

∴ LHS = RHS (verified)

j) 2q – 6 = 0

Solution:

The given equation is 2q – 6 = 0

Add 6 to both the sides

Substitute the value of q = 3 in the given equation

2q – 6 = 0

2(3) – 6 = 0

6 – 6 = 0

0 = 0

LHS = RHS (verified)

k) 2q + 6 = 0

Solution:

The given equation is 2q + 6 = 0

Subtract 6 from both the sides

Divide by 2 both the sides by 2

Substitute the value of q = -3 in the given equation

2q + 6 = 0

2(-3) + 6 = 0

-6 + 6 = 0

0 = 0

∴ LHS = RHS (verified)

l) 2q + 6 = 12

The given equation is 2q + 6 = 12

Subtract 6 from both the sides

Divide by 2 both the sides