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NCERT solution class 7 chapter 4 simple equations exercise 4.2 mathematics

EXERCISE 4.2


QUESTION 1

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

(a) x+ 1 = 0

Sol :

Adding 1 to both sides of the given equation, we obtain

x – 1 + 1 = 0 + 1

x = 1

 

(b) x + 1 = 0

Sol :

Subtracting 1 from both sides of the given equation , we obtain

x + 1 – 1 = 0 – 1

x = – 1

 

(c) x – 1 = 5

Sol :

Adding 1 to both sides of the given equation, we obtain

x – 1 + 1 = 5 + 1

x = 6

 

(d) x + 6 = 2

Sol :

Subtracting 6 from both sides of the given equation, we obtain

x + 6 – 6 = 2 – 6

x = – 4

 

(e) y – 4 = -7

Sol :

adding 4 to both sides of the given equation , we obtain

y – 4 + 4= -7 + 4

y = -3

 

(f) y – 4 = 4

Sol :

Adding 4 to both sides of the given equation, we obtain

y – 4 + 4 = 4 + 4

y = 8

 

(g) y + 4 = 4

Sol :

Subtracting 4 from both sides of the given equation , we obtain

y + 4 – 4 = 4 – 4

y = 0

 

(h) y + 4 = -4

Sol :

Subtracting 4 from both sides of the given equation , we obtain

y + 4 – 4 = 4 – 4

y = -8

 


QUESTION 2

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

(a) 3l = 42

Sol :

Dividing both sides of the given equation by 3, we obtain

\dfrac{3l}{3}=\dfrac{42}{3}

l = 14

 

(b) \dfrac{b}{2}=6

Sol :

Multiplying both sides of the given equation by 2, we obtain

\dfrac{b\times 2}{2}=6\times 2

b = 12

 

(c) \dfrac{p}{7}=4

Sol :

Multiplying both sides of the given equation by 7, we obtain

\dfrac{p\times 7}{7}=4\times 7

p = 28

 

(d) 4x = 25

Sol :

Dividing both sides of the given equation by 4, we obtain

\dfrac{4x}{4}=\dfrac{25}{4}

x=\dfrac{25}{4}

 

(e) 8y = 36

Sol :

Dividing both sides of the given equation by 8, we obtain

\dfrac{8y}{8}=\dfrac{36}{8}

y=\dfrac{9}{2}

 

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

Sol :

Multiplying both sides of the given equation by 3, we obtain

\dfrac{z\times 3}{3}=\dfrac{5\times 3}{4}

z=\dfrac{15}{4}

 

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

Sol :

Multiplying both sides of the given equation by 5, we obtain

\dfrac{a\times 5}{5}=\dfrac{7\times 5}{15}

a=\dfrac{7}{3}

 

(h) 20t = -10

Sol :

Dividing both sides of the given equation by 20, we obtain

\dfrac{20t}{20}=\dfrac{-10}{\phantom{-}20}

t=\dfrac{-1}{\phantom{-}2}

 


QUESTION 3

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

(a) 3n – 2 = 26

Sol :

Adding 2 to both sides of the given equation, we obtain

3n – 2 + 2 = 46 + 2

3n = 48

Dividing both sides of the given equation by 3, we obtain

\dfrac{3n}{3}=\dfrac{48}{3}

n = 16

 

(b) 5m + 7 = 7

Sol :

Subtracting 7 from both sides of the given equation, we obtain

5m + 7 – 7 = 17 – 7

5m = 10

Dividing both sides of the given equation by 5, we obtain

\dfrac{5}{m}=\dfrac{10}{5}

m = 2

 

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

Sol :

Multiplying both sides of the given equation by 3, we obtain

\dfrac{20p\times 3}{3}=40\times 3

20p = 120

Dividing both sides of the given equation by 20, we obtain

\dfrac{20p}{20}=\dfrac{120}{20}

p = 6

 

(d) \dfrac{3p}{10}=6

Sol :

Multiplying both sides of the given equation by 10, we obtain

\dfrac{3p\times 10}{10}=6\times 10

3p = 60

Dividing both sides of the given equation by 3, we obtain

\dfrac{3p}{3}=\dfrac{60}{3}

p = 20

 


QUESTION 4

Solve the following equations:

(a) 10p = 100

Sol :

\dfrac{10p}{10}=\dfrac{100}{10}

p = 10

 

(b) 10p + 10

Sol :

10p + 10 – 10 = 100 – 10

10 p = 90

\dfrac{10p}{10}=\dfrac{90}{10}

p = 9

 

(c) \dfrac{p}{4}=5

Sol :

\dfrac{p\times 4}{4}=5\times 4

p = 20

 

(d) \dfrac{-p}{\phantom{-}3}=5

Sol :

\dfrac{-p\times (-3)}{\phantom{-}3}=5\times (-3)

p = -15

 

(e) \dfrac{3p}{4}=6

Sol :

\dfrac{3p\times 4}{4}=6\times 4

3p = 24

p = 8

 

(f) 3s = -9

Sol :

\dfrac{3s}{3}=\dfrac{-9}{\phantom{-}3}

x = -3

 

(g) 3s + 12  = 0

Sol :

3s + 12 -12 = 0 -12

3s = -12

\dfrac{3s}{3}=\dfrac{-12}{\phantom{-}3}

s = -4

 

(h) 3s = 0

Sol :

\dfrac{3s}{3}=\dfrac{0}{3}

s = 0

 

(i) 2q = 6

Sol :

\dfrac{2q}{2}=\dfrac{6}{2}

q = 3

 

(j) 2q – 6 = 0

Sol :

2q – 6 + 6 = 0 + 6

2q = 6

\dfrac{2q}{2}=\dfrac{6}{2}

q = 3

 

(k) 2q + 6 = 0

Sol :

2q + 6 – 6 = 0

2q = -6

\dfrac{2q}{2}=\dfrac{-6}{\phantom{-}2}

q = -3

 

(l) 2q + 6 = 12

Sol :

2q + 6 – 6 = 12 – 6

2q = 6

\dfrac{2q}{2}=\dfrac{6}{2}

q = 3


 

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