NCERT MATHS SOLUTION CLASS 7 EXERCISE 4.3 | SIMPLE EQUATIONS

Find here step-by-step NCERT Maths Solution Class 7 Exercise 4.3 | Solving Equations | Chapter 4 Simple Equations from CBSE NCERT textbook.

Before moving to maths solution class 7 exercise 4.3 , it is advised to go through the following topics:
• Simple Equations

NCERT Maths Solution Class 7 Exercise 4.3

Problems in Exercise 4.3 require the knowledge of finding the Solution of a Simple Equation, Transposing etc.

The properties of equality gives a systematic method of solving an equation. We carry out a series of identical mathematical operations on the two sides of the equation in such a way that on one of the sides we get just the variable. The last step is the solution of the equation.

Transposing means moving to the other side. Transposition of a number has the same effect as adding same number to (or subtracting the same number from) both sides of the equation. When we transpose a number from one side of the equation to the other side, we change its sign.

Exercise 4.3
Solving Simple Equations | Transposing

QUESTION 1

Solve the following equations:

(a) 2y + \frac{5}{2} = \frac{37}{2}

(b) 5t + 28 = 10

(c) \frac{a}{5} + 3 = 2

(d) \frac{q}{4} + 7 = 5

(e) \frac{5}{2}\times x = -5

(f) \frac{5}{2}\times x = \frac{25}{4}

(g) 7m + \frac{19}{2} = 13

(h) 6z + 10 = –2

(i) \frac{3l}{2} = \frac{2}{3}

(j) \frac{2b}{3} - 5 = 3

ANSWER

(a) 2y + \frac{5}{2} = \frac{37}{2}

\implies 2y = \frac{37}{2}-\frac{5}{2}

\implies 2y = \frac{37-5}{2}

\implies 2y = \frac{32}{2}

\implies y = \frac{16}{2}

\implies y = 8

(b) 5t + 28 = 10

=> 5t = 10 – 28

=> t = – 18/5

=> t = – 3.6

(c) \frac{a}{5} + 3 = 2

\implies \frac{a}{5}=2-3

\implies a = -1\times 5

\implies a = -5

(d) \frac{q}{4} + 7 = 5

\implies \frac{q}{4}=5-7

\implies q = -2\times 4

\implies q = -8

(e) \frac{5}{2}\times x = -5

\implies x = -5\times \frac{2}{5}

\implies x = -2

(f) \frac{5}{2}\times x = \frac{25}{4}

\implies x = \frac{25}{4}\times \frac{2}{5}

\implies x = \frac{5}{2}

\implies x = 2.5

(g) 7m + \frac{19}{2} = 13

\implies 7m = 13-\frac{19}{2}

\implies 7m = \frac{26-19}{2}

\implies 7m = \frac{7}{2}

\implies m = \frac{7}{2\times 7}

\implies m = 0.5

(h) 6z + 10 = –2

\implies 6z = -2-10

\implies z = \frac{-12}{6}

\implies z = -2

(i) \frac{3l}{2} = \frac{2}{3}

\implies 3l = \frac{2}{3}\times 2

\implies l = \frac{4}{3\times 3}

\implies l = \frac{4}{9}

(j) \frac{2b}{3} - 5 = 3

\implies \frac{2b}{3}=3+5

\implies 2b = 8\times 3

\implies b = \frac{24}{2}

\implies b = 12

[ NCERT Maths Solution Class 7 Exercise 4.3 ]

QUESTION 2

Solve the following equations:

(a) 2(x + 4) = 12 (b) 3(n – 5) = 21

(c) 3(n – 5) = – 21 (d) – 4(2 + x) = 8

(e) 4(2 – x) = 8

ANSWER

(a) 2(x + 4) = 12

=> x + 4 = \frac{12}{2}

=> x = 6 – 4

=> x = 2

(b) 3(n – 5) = 21

=> n – 5 = \frac{21}{3}

=> n = 7 + 5

=> n = 12

(c) 3(n – 5) = – 21

=> n – 5 = \frac{-21}{3}

=> n = – 7 + 5

=> n = – 2

(d) – 4(2 + x) = 8

=> 2 + x = \frac{8}{-4}

=> x = – 2 – 2

=> x = – 4

(e) 4(2 – x) = 8

=> 2 – x = \frac{8}{4}

=> – x = 2 – 2

=> x = 0

[ NCERT Maths Solution Class 7 Exercise 4.3 ]

QUESTION 3

Solve the following equations:

(a) 4 = 5(p – 2)

(b) – 4 = 5(p – 2)

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

(d) 4 + 5(p – 1) = 34

(e) 0 = 16 + 4(m – 6)

ANSWER

(a) 4 = 5(p – 2)

=> 5(p – 2) = 4

=> p – 2 = \frac{4}{5}

=> p = \frac{4}{5} + 2

=> p = \frac{4+10}{5}

=> p = \frac{14}{5}

=> p = 2.8

(b) – 4 = 5(p – 2)

=> 5(p – 2) = – 4

=> p – 2 = \frac{-4}{5}

=> p = \frac{-4}{5} + 2

=> p = \frac{-4+10}{5}

=> p = \frac{6}{5}

=> p = 1.2

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

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

=> 3(t + 2) = 16 – 4

=> (t + 2) = \frac{12}{3}

=> t = 4 – 2

=> t = 2

(d) 4 + 5(p – 1) = 34

=> 5(p – 1) = 34 – 4

=> (p – 1) = \frac{30}{5}

=> p = 6 + 1

=> p = 7

(e) 0 = 16 + 4(m – 6)

=> 16 + 4(m – 6) = 0

=> 4(m – 6) = – 16

=> (m – 6) = \frac{-16}{4}

=> m = – 4 + 6

=> m = 2

[ NCERT Maths Solution Class 7 Exercise 4.3 ]

QUESTION 4

(a) Construct 3 equations starting with x = 2

(b) Construct 3 equations starting with x = – 2

ANSWER

(a) x = 2

Multiplying both sides by 3

3x = 6 … (i)

Subtracting one from both sides

3x – 1 = 5 … (ii)

Dividing both sides by 3

x - \frac{1}{3}=\frac{5}{3} … (iii)

(b) x = – 2

Adding 5 to both sides

x + 5 = 3 … (i)

Multiplying both sides by 2

2x + 10 = 6 … (ii)

Subtracting 10 from both sides

2x = – 4 … (iii)

NCERT MATHS SOLUTION | CLASS 7 | CHAPTER 4 | SIMPLE EQUATIONS

Exercise 4.1 | Exercise 4.2

Exercise 4.3 | Exercise 4.4

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