HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.3

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 4 Simple Equations Exercise 4.3

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) \(\frac{5}{2}\)x = -10
(f) \(\frac{5}{2} x=\frac{25}{4}\)
(g) 7m + \(\frac{19}{2}\) = 13
(h) 6z + 10 = -2
(i) \(\frac{3 \cdots}{2}=\frac{2}{3}\)
(j) \(\frac{2b}{3}\) – 5 = 3
Solution:
(a) 2y + \(\frac{5}{2}=\frac{37}{2}\)
\(2 y=\frac{37-5}{2}=\frac{32}{2}=\frac{37}{2}-\frac{5}{2}\)
2y = 16
y = \(\frac{16}{2}\)
∴ y = 8

HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.3

(b) 5t + 28 = 10
or, 5t = 10 – 28 = -18
∴ t = \(\frac{-18}{5}\)

(c) \(\frac{a}{5}\) + 3 = 2
or \(\frac{a}{5}\) = 1 – 3 = -1
a = -1 × 5 = -5
∴ a = -5

(d) \(\frac{a}{4}\) + 7 = 5
or, \(\frac{a}{4}\) = 5 – 7 = -2
a = -2 × 4 = -8
∴ a = -8

(e) \(\frac{5}{2}\)x = -10
5x = 2 × (-10) = -20
x = \(\frac{-20}{5}\)
∴ x = -4

(f) \(\frac{5}{2} x=\frac{25}{4}\)
5x × 4 = 25 × 2
20x = 50
x = \(\frac{50}{20}=\frac{5}{2}\)
∴ x = \(\frac{5}{2}\)

(g) 7m + \(\frac{19}{2}\) = 13
or, 7m = 13 – \(\frac{19}{2}\)
= \(\frac{29-19}{2}=\frac{7}{2}\)
or, 7m × 2 = 7
or, m = \(\frac{7}{7 \times 2}=\frac{1}{2}\)
∴ m = \(\frac{1}{2}\)

(h) 6z + 10 = -2
or, 6z = -2 – 10 = -12
or z = \(\frac{-12}{6}\) = -2
∴ z = -2

HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.3

(i) \(\frac{3 \cdots}{2}=\frac{2}{3}\)
3l × 3 = 2 × 2 = 4
∴ l = \(\frac{4}{9}\)

(j) \(\frac{2b}{3}\) – 5 = 3
\(\frac{2b}{3}\) = 3 + 5 = 8
2b = 3 × 8 = 24
b = \(\frac{24}{2}\) = 12
∴ b = 12

Question 2.
Solve the following equations :
(a) 2(x + 4) = 12
(b) 3(n-5)=21
(c) 3(n-5) = -21
(d) 3 – 2(2 – y)-7
(e) – 4(2 – x) = 9
(f) 4(2 – x) = 9
(g) 4 + 5 (p – 1) = 34
(h) 34 – 5(p – 1) = 4
Solution:
(a) 2(x + 4) = 12
or, x + 4 = \(\frac{12}{2}\) = 6
or, x = 6 – 4 = 2
∴ x = 2

(b) 3(n – 5) = 21
or, (n – 5) = \(\frac{21}{3}\) = 7
n = 7 + 5 = 12
∴ n = 12

(c) 3(n – 5) = -21
or, (n – 5) = \(\frac{-21}{3}\)= – 7
n = – 7 + 5 = – 2
∴ n = -2

(d) 3 – 2(2 – y) = 7
or, -2(2 – y) = 7 – 3 = 4
or, (2 – y) = \(\frac{4}{-2}\) = -2
-y = – 2 – 2 = – 4
∴ y = 4

(e) -4(2 – x) = 9
or, 2-x = \(\frac{9}{-4}\)
or, -x = \(\frac{-9-8}{4}=\frac{-17}{4}\)
∴ x = \(\frac{17}{4}\)

(f) 4(2 – x) = 9
or, 2 – x = \(\frac{9}{4}\)
or – x = \(\frac{9}{4}\) – 2
= \(\frac{9-8}{4}=\frac{1}{4}\)
∴ x = \(-\frac{1}{4}\)

HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.3

(h) 34 – 5(p – 1) = 4
or, – 5(p – 1) = 4 – 34 = – 30
or, (p-1) = \(\frac{-30}{-5}\) = 6
or, P = 6 + 1
∴  P = 7.

Question 3.
Solve the following equations :
(a) 4 = 5(p-2)
(b) – 4 = 5(p – 2)
(c) -16 = -5(2 – p)
(d) 10 = 4 + 3(t + 2)
(e) 28 = 4 + 3(t + 5)
(f) 0 = 16 + 4(m -6)
Solution:
(a) 4 = 6(p – 2)
or, 5(p – 2) = 4
or, (p – 2) = \(\frac{4}{5}\)
or, P = \(\frac{4}{5}+2=\frac{4+10}{5}=\frac{14}{5}\)
∴ P = \(\frac{14}{5}\)

(b) -4 = 5(p – 2)
or, 5 (p – 2) = _4
(p-2) = \(\frac{-4}{5}\)
p = \(\frac{-4}{5}+2=\frac{-4+10}{5}=\frac{6}{5}\)
∴ p = \(\frac{6}{5}\)

(c) -16 = -5(2-p)
or, 5(2-p) = 16
2-p = \(\frac{16}{5}\)
-p = \(\frac{16}{5}-2=\frac{16-10}{5}=\frac{6}{5}\)
∴ p = \(\frac{6}{5}\)

(d) 10 = 4 + 3(t + 2)
or, 4 + 3(t + 2) = 10
3 (t + 2) = 10 – 4 = 6
t + 2 = \(\frac{6}{3}\) = 2
t = 2 – 2 = 0
∴ t = 0

(e) 28 = 4 + 3(t + 5)
or, 4 + 3(t + 5) = 28
3(t + 5) = 28-4 = 24
(t + 5) = \(\frac{24}{3}\) = 8
t = 8-5 = 3
∴ t = 3

(f) 0 = 16 + 4(m – 6)
or, 16 + 4(m – 6) = 0
or, 4 (m -6) = -16
(m – 6) = \(\frac{-16}{4}\) = – 4
m = -4 + 6 = 2
∴ m = 2.

HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.3

Question 4.
(a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x=- 2.
Solution:
(a) x = 2
2x = 2 × 2 = 4 ⇒ 2x = 4
3x = 2 × 3 = 6 ⇒ 3x= 6
4x = 2 × 4 = 8 ⇒ 4x = 8

(b) x =-2
2x = -2 × 2 = -4 ⇒ 2x = -4
3x = – 2 × 3 = – 6 ⇒ 3x = – 6
5x = – 2 × 5 = – 10 ⇒ 5x = -10.

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