**Simple Equations Class 7 Math EX-4.1**

**Simple Equations for Class 7, Chapter 4 Maths NCERT Solution ;**

**Simple Equations Class 7 Ex. 4.2;**

**Exercise 4.2**

**Simple Equations Chapter 4 NCERT Math Ex.-4.2**

**Question 1:- Gives, the first the step you will use to separate the variable and then solve the equation:**

**(a) x – 1 = 0**

**(b) x + 1 = 0**

**(c) x – 1 = 5**

**(d) x + 6 = 2**

**(e) y – 4 = -7**

**(f) y – 4 = 4**

**(g) y + 4 = 4**

**(h) y + 4 = – 4**

**Solution 1:-**

(a) x – 1 = 0

Adding ”1” to both sides, we get

x – 1 + 1 = 0 + 1

x = 1

Therefore, the necessary solution is x = 1.

**Verify:** Enter x = 1 in the provided equations.

x – 1 = 0

1 – 1 = 0

0 = 0

L.H.S = R.H.S

Thus x = 1 is required solution.

(b) x + 1 = 0

Subtracting 1 from both sides, we get

x + 1 – 1 = 0 – 1

x = -1

Therefore, the necessary solution is x = -1.

**Verify:** Enter x = -1 in the provided equations.

x+1=0

-1 + 1 = 0

0 = 0

L.H.S = R.H.S

Thus, the needed solution is x = -1.

(c) x – 1 = 5

Adding 1 to both sides, we get

x – 1 + 1 = 5 + 1

x = 6

Therefore, the necessary solution is x = 6.

**Verify:** Enter x = 6 in the provided equations.

x – 1 = 5

6 – 1 = 5

5 = 5

L.H.S = R.H.S

Thus, the right answer is x = 6.

(d) x + 6 = 2

Subtracting 6 from both sides, we get

x + 6 – 6 = 2 – 6

x = -4

Therefore, the necessary solution is x = -4.

**Verify:** Enter x = -4 in the provided equations.

x + 6 = 2

Putting x = -4, we get

-4 + 6 = 2

2 = 2

L.H.S = R.H.S

Therefore, the necessary solution is x = -4.

(e) y – 4 = -7

Adding 4 to both sides, we get

y – 4 + 4 = -7 + 4

y = -3

Therefore, the necessary solution is y = -3.

**Verify:** Enter y = -3 in the provided equations.

y – 4 = -7

Putting y = -3, we get

-3 – 4 = -7

⇒ -7 = -7

L.H.S = R.H.S

Consequently, the necessary solution is y = -3.

(f) y – 4 = 4

Adding 4 to both sides, we get

y – 4 + 4 = 4 + 4

y = 8

Therefore, the necessary solution is y = 8.

**Verify:** Enter y = 8 in the provided equations.

y – 4 = 4

8 – 4 = 4

4 = 4

L.H.S = R.H.S

Thus, the right answer is y = 8.

(g) y + 4 = 4

Subtracting 4 from both sides, we get

y + 4 – 4 = 4 – 4

y = 0

Therefore, the necessary solution is y = 0.

**Verify:** Enter y = 0 in the provided equations.

y + 4 = 4

0 + 4 = 4

4 = 4

L.H.S = R.H.S

Therefore, y = 0 is the required solution.

(h) y + 4 =-4

Subtracting 4 from both sides, we get

y + 4 – 4 = -4 – 4

y = -8

Therefore, the necessary solution is y = -8.

**Verify:** Enter y = -8 in the provided equations.

y + 4 = -4

Putting y = -8, we get

-8 + 4 = -4

-4 = -4

L.H.S = R.H.S

Therefore, y = -8 is the required solution.

**Simple Equations Chapter 4 NCERT Math Ex.-4.2**

**Question 2 :- Give first the step you will use to separate the variable and then solve the following equation:**

**(a) 3L = 42**

**(b)b/2=6**

**(c)p/7=4**

**(d)4x=25**

**(e)8y=36**

**(f)z/3=5/4**

**(g)a/5=7/15**

**(h)20t=-10**

**Solution 2:-**

(a) 3L = 42

Dividing both side by 3, we get

3L/3 = 42/3

L = 14

Thus, L = 14 is the required solution.

**Check:** 3L = 42

Putting y = 14, we get

3×14 = 42 ⇒ 42 = 42

L.H.S = R.H.S

Thus, 14 = 14 is the correct solution.

(b) b/2 = 6

Multiply both side by 2, we get

(b/2)x2 = 6×2

b = 12

Thus, b = 12 is the required solution.

**Check:** b = 12

Putting b = 14, we get

3×14 = 42 ⇒ 42 = 42

L.H.S = R.H.S

Thus, 14 = 14 is the correct solution.

(c) p/7 = 4

Multiply both side by 7, we get

(p/7)x7 = 4×7

p = 28

Thus, p = 28 is the required solution.

**Check:** p/7 = 4

Putting p = 28, we get

28/7 = 4 ⇒ 4 = 4

L.H.S = R.H.S

Thus, 4 = 4 is the correct solution.

(d) 4x = 25

Dividing both side by 4, we get

(4x)/4 = 25

x = 25

Thus, x = 25 is the required solution.

**Check:** 4x = 25

Putting x = 25, we get

25 = 25 ⇒ 25 = 25

L.H.S = R.H.S

Thus, 25 = 25 is the correct solution.

(e) 8y = 36

Dividing both side by 8, we get

(8y)/8 = 36/8

y = 9/2

Thus, y = 9/2 is the required solution.

**Check:** 8y = 36

Putting y = 9/2, we get

8x(9/2) = 36 ⇒ 36 = 36

L.H.S = R.H.S

Thus, 36 = 36 is the correct solution.

(f) z/3 = 5/4

Multiply both side by 3, we get

(z/3)x3 = (5/4)x(3)

z = 15/4

Thus, z = 15/4 is the required solution.

**Check:** z/3 = 5/4

Putting z = 15/4, we get

(15/4)x(1/3) = 5/4

5/4 = 5/4

L.H.S = R.H.S

Thus, 5/4 = 5/4 is the correct solution.

(g) a/5 = 7/15

Multiply both side by 5, we get

(a/5)x5 = (7/15)x5

a = 7/3

Thus, a = 7/15 is the required solution.

**Check:** a/5 = 7/15

Putting a = 7/3, we get

(7/3)x(1/5) = 7/15 ⇒ 7/15 = 7/15

L.H.S = R.H.S

Thus, 7/15 = 7/15 is the correct solution.

(h) 20t = -10

Dividing both side by 20, we get

(20t)/20 = -(10/20)

t = -(1/2)

Thus, t = -(1/2) is the required solution.

**Check:** 20t = -10

Putting t = -(1/2), we get

20x-(1/2) = -10 ⇒ -10 = -10

L.H.S = R.H.S

Thus, -10 = -10 is the correct solution.

**Simple Equations Chapter 4 NCERT Math Ex.-4.2**

**Question 3 :- Gives, the steps you will use to separate the variables and then solve the equation:**

**(a) 3n – 2 = 46**

**(b) 5m + 7 = 17**

**(c) 20p/3=40**

**(d) 3p/10=6**

**Solution 3:-**

(a) 3n – 2 = 46

⇒ 3n – 2 + 2 = 46+ 2 [Since, adding 2 to both sides]

⇒ 3n = 48

⇒ 3n ÷ 3 = 48 ÷ 3 [Since, Dividing both sides by 3]

n = 16

**Check:** 3n-2 = 46

Putting n = 16, we get

3x(16) – 2 = 46

48-2 = 46

46=46

L.H.S = R.H.S

Thus, 46 = 46 is the correct solution.

(b) 5m + 7 = 17

⇒ 5m + 7 – 7 = 17- 7 [Since, Subtracting 7 to both sides]

⇒ 5m = 10

⇒ 5m/5 = 10/5 [Since, Dividing both sides by 5]

m = 2

**Check:** 5m+7 = 17

Putting m = 2, we get

5x(2) + 7 = 17

10+7 = 17

17=17

L.H.S = R.H.S

Thus, 17 = 17 is the correct solution.

(c) 20p/3 = 40

⇒ (20p/3)x3 = 40×3 [Since, Multiplying 3 to both sides]

⇒ 20p = 120

⇒ 20p/20 = 120/20 [Since, Dividing both sides by 20]

p = 6

**Check:** 20p/3 = 40

Putting p = 6, we get

20x(6)/3 = 40

40 = 40

40=40

L.H.S = R.H.S

Thus, 40 = 40 is the correct solution.

(d) 3p/10 = 6

⇒ (3p/10)x10 = 6×10 [Since, Multiplying 10 to both sides]

⇒ 3p = 60

⇒ 3p/3 = 60/3 [Since, Dividing both sides by 3]

p = 20

**Check:** 3p/10 = 6

Putting p = 20, we get

3x(20)/10 = 6

60/10 = 6

6=6

L.H.S = R.H.S

Thus, 6 = 6 is the correct solution.

**Simple Equations Chapter 4 NCERT Math Ex.-4.2**

**Question 4:- Solve the following equations:**

**(a) 10p = 100**

**(b) 10p + 10 = 100**

**(c) p/4=5**

**(d) −p/3=5**

**(e) 3p/4=6**

**(f) 3s = -9**

**(g) 3s + 12 = 0**

**(h) 3s = 0**

**(i) 2q = 6**

**(j) 2q – 6 = 0**

**(k) 2q + 6 = 0**

**(l) 2q + 6 = 12**

**Solution 4:-**

(a) 10p = 100

⇒ 10p/10 = 100/10 [Since, Dividing both sides by 10]

p=100/10=10

Thus p= 10

(b) 10p + 10 = 100

⇒ 10p + 10 – 10 = 100 -10 [Since, Subtracting 10 from both sides]

⇒ 10p = 90

⇒ 10p/10 = 90/10 [Since, Dividing both side by 10]

p=90/10=9

Thus p = 9

(c) p/4 = 5

⇒ (p/4)x(4) = 5×4 [Since, Multiplying 4 from both sides]

⇒ p = 20

(d) -p/3 = 5

⇒ -(p/3)x(3) = 5×3 [Since, Multiplying 3 from both sides]

⇒ p = -15

(e) 3p/4 = 6

⇒ (3p/4)x(4) = 6×4 [Since, Multiplying 4 from both sides]

⇒ 3p = 24 [Dividing both side by 3]

3p/3 = 24/3

p=8

(f) 3s = -9

⇒ (3s)/(3) = -9/3 [Since, Dividing 3 from both sides]

⇒ s = -3

(g) 3s+12 = 0

⇒ 3s+12-12 = 0-12 [Since, Subtracting 12 from both sides]

⇒ 3s = -12 [Dividing both side by 3]

3s/3 = -12/3

s=-4

(h) 3s = 0

⇒ 3s/3 = 0/3 [Since, Dividing 3 from both sides]

⇒ s = 0

(i) 2q = 6

⇒ (2q )/(2) = 6/2 [Since, Dividing 2 from both sides]

⇒ q = 3

(j) 2q-6 = 0

⇒ 2q -6+6 = 0+6 [Since, Adding 6 from both sides]

⇒ 2q = 6 [Dividing both side by 2]

2q/2=6/2

q=3

(k) 2q+6 = 0

⇒ 2q +6-6 = 0-6 [Since, Subtracting 6 from both sides]

⇒ 2q = -6

2q/2=-6/2 [Dividing both side by 2]

q=-3

(l) 2q + 6 = 12

⇒ 2q + 6 – 6 = 12 – 6 [Since, Subtracting 6 from both sides]

⇒ 2q = 6

2q/2=6/2 [Dividing both side by 2]

q=3

**Simple Equations Chapter 4 NCERT Math Ex-4.2**

Simple Equations

Simple Equations

Simple Equations

Simple Equations

Simple Equations

Simple equations