Students can Download Chapter 4 Simple Equations Ex 4.3, 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 4.3

Question 1.

Solve the following equations :

Solution:

Substitute the value of y = 8 in the given equation

∴ LHS = RHS (verified)

b) 5t + 28 = 10

Solution:

The given equation is 5t + 28 = 10

Transposing 28 from LHS to RHS

5t = 10 – 28 [on transposing + 28 becomes -28]

5t = -18

Dividing by 5 both the sides

Substitute the value of t = \(\frac{-18}{5}\) in the given equation

5t + 28 = 10

-18 + 28 = 10

10 = 10

∴ LHS = RHS (verified)

c)

Solution:

Transposing + 3 from LHS to RHS

Multiply by 5 both the sides

Substitute the value of a = -5 in the given equation.

-1 + 3 = 2

2 = 2

∴ LHS = RHS (verified)

d)

Solution:

Transposing +7 from LHS to RHS

Substitute the value of q = -8 in the given equation.

-2 + 7 = 5

5 = 5

∴ LHS = RHS (verified)

e)

Solution:

Multiply by 2 both the sides

Substitute the value of x = -4 in the given equation.

∴ LHS = RHS (verified)

f)

Solution:

Multiply by 2 both the sides

Divide by 5 both the sides

∴ LHS = RHS (verified)

g)

Solution:

Divide by 7 both the sides

Substitute the value of m = \(\frac{1}{2}\) in the given equation

13 = 13

∴ LHS = RHS (verified)

h) 6z + 10 = -2

Solution:

The given equation is 6z + 10 = -2

Transposing + 10 from LHS to RHS

6z = -2 – 10(on Transposing + 10 becomes – 10)

6z = -12

Divide by 6 both the sides

Substitute the value of z = -2 in the given equation

6z + 10 = -2

6(-2) + 10 = -2

-12 + 10 = -2

-2 = -2

∴ LHS = RHS (verified)

i)

Solution:

multiply by 2 both the sides

Divide by 3 both the sides

Substitute the value of l = \(4 / 9\) in the given equation.

∴ LHS = RHS (verified)

j)

Solution:

Transpose -5 from LHS to RHS

Multiply by 3 both the sides

Divide by 2 both the sides

Substitute the value of b = 12 in the given equation

3 = 3

∴ LHS = RHS (verified)

Question 2.

Solve the Mowing equations :

a) 2(x+4) = 12

Solution:

The given equation is 2(x + 4) = 12

Divide by 2 both the sides

x + 4 = 6

Transpose +4 from LHS to RHS x = 6 – 4 (on transposing 4 becomes -4)

∴ x = 2

Substitute the value of x = 2 in the given equation

2(x + 4) = 12

2(2 + 4) = 12

2(6) = 12

12 = 12

∴ LHS = RHS (verified)

b) 3(n – 5) = 21

The given equation is 3 (n – 5) = 21

Divide by 3 both the sides

∴ n – 5 = 7

Transpose -5 from LHS to RHS

n = 7 + 5 [on transposing -5 becomes +5]

∴ n = 12

Substitute the value of n = 12 in the given equation

3(n – 5) = 21

3(12 – 5) = 21

3(7) = 21

21 = 21

LHS = RHS (verified)

c) 3 (n – 5) = -21

The given equation is 3 (n – 5) = -21

Divide by 3 both the sides 3

Transpose -5 from LHS to RHS n = -7 + 5 (on transposing -5 becomes +5)

n = -2

Substitute the value n = -2 in the given equation

3 (n – 5) = -21 .

3(-2 – 5) = -21

3(-7) = -21

-21 = -21

∴ LHS = RHS (verified)

d) -4(2 + x) = 8

Solution:

The given equation is -4 (2 + x) = 8

Divide by -4 both the sides

2 + x = -2

Transpose 2 from LHS to RHS x = -2 -2 [on transposing 2 becomes -2]

∴ x = -4

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

-4(2 + x) = 8

-4(2 – 4) = 8

-4 (-2) = 8

8 = 8

∴ LHS = RHS (verified)

e) 4 (2 – x) = 8

The given equation is 4 (2 – x) = 8

Divide by 4 both the sides

2 – x = 2

Transpose +2 from LHS to RHS

-x = 2 – 2 (on transposing 2 becomes -2)

-x = 0

multiply by -1 both the sides

-x × -1 = 0 × -1 x = 0

Substituting the value of x = 0 in the given equation

4(2 – x) = 8

4(2 – 0) = 8

8 – 0 = 8

8 = 8

∴ LHS = RHS (verified)

Question 3.

Solve the following equations :

a) 4 = 5 (p – 2)

Solution:

The given equation is 4 = 5 (p – 2)

5(p – 2) = 4 [LHS and RHS are inter-changed the equation remains the same]

Divide by 5 both sides

Transpose -2 from LHS to RHS

Substituting the value of p = \(\frac{14}{5}\) in the given equation

∴ LHS = RHS (verified)

b) -4 = 5 (p – 2)

Solution:

The given equation is – 4 = 5 (p – 2)

[LHS & RHS are interchanged the equation remains same]

5 (p – 2) = -4

Divide by 5 both the sides

Transpose -2 from LHS to RHS

Substitute the value of P = 6/5 in the given equation

5(p – 2) = -4

∴ LHS = RHS (verified)

c) 16 = 4 + 3 (t + 2)

Solution:

The given equation is 16 = 4 + 3 (t + 2)

4 + 3 (t + 2) = 16 [LHS and RHS are interchanged the equation remains the same.] 4 + 3t + 6 = 16

3t + 10 = 16

3t = 16 – 10 [ transposing + 10 from LHS to RHS, it becomes -10]

3t = 6

t = 6/3 = 2 ∴ t = 2

Substitute the value of t = 2 in the given equation

16 = 4 + 3 (t +2)

16 = 4 + 3t + 6

16 = 10 + 3(2)

16 = 10 + 6

16 = 16

∴ LHS = RHS (verified)

d) 4 + 5 (p – 1) = 34

Solution:

The given equation is 4 + 5 (p – 1) = 34

Transpose 4 from LHS to RHS

5(p – 1) = 34 – 4 (on transposing 4 becomes -4)

5(p – 1) = 30

Divide by 5 both sides

p – 1 = 6

Transposing -1 from LHS to RHS p = 6 + 1 [on transposing -1 becomes 1]

p = 7

Substitute the value of p = 7 in the given equation

4 + 5(p – 1) = 34

4 + 5p – 5 = 34

4 + 5 (7) – 5 = 34

4 + 35 – 5 = 34

34 = 34

∴ LHS = RHS (verified)

e) 0 = 16 + 4(m – 6)

Solution:

The given equation is 0 = 16 + 4 (m – 6)

16 + 4 (m – 6) = 0 [ LHS ami RHS are interchanged the equation remains the same] Transpose 16 from LHS to RHS .

4(m – 6) = 0 – 16 [on transposing 16 becomes -16]

4(m – 6) = -16

Divide by 4 both sides

m – 6 = – 4

Transpose – 6 from LHS to RHS

m = – 4 + 6 [on transposing – 6 becomes + 6]

m = 2

Substitute the value m = 2 in the given equation

16 + 4(m – 6) = 0

16 + 4(2 – 6) = 0

16 + 4(-4) = 0

16 – 16 = 0

0 = 0

∴ LHS = RHS (verified)

Question 4.

a) Construct 3 equations starting with x = 2

Solution:

x = 2

multiply by 3 both sides

3(x) = 2(3)

3x = 6

Add 6 to both sides

3x + 6 = 6 + 6

equation 1 3x + 6 = 12

x = 2

multiply by 12 both sides

12(x) = 2(12)

12x = 24

Subtract 10 from both sides

12x – 10 = 24 – 10

equation 2 12x – 10 = 14

multiply by 15 both the sides

x (15) = 2(15)

15x = 30

Divide by 2 both the sides

b) Construct 3 equations starting with x = -2

Solution:

x = -2

multiply by 5 both sides

5(x) = -2 (5)

5x = -10

x = -2

multiply by 8 both the sides

8(x) = -2 × 8

8x = -16

Divide by 3 both the sides

x = -2

multiply by 9 both the sides

9(x) = -2 × 9

9x = -18

Subtract 7 from both the sides 9x-l – -18 – 7

3. 9x – 7 = -25 —– 3. Equation