RBSE Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2

Rajasthan Board RBSE Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Textbook Exercise Questions and Answers.

RBSE Class 7 Maths Solutions Chapter 4 Simple Equations Ex 4.2

Question 1.
Give first the .step you will use to separate the variable and then solve the equation:
(a) x - 1 = 0
Answer:
Given equation is x - 1 = 0
(Adding 1 to both sides) .
x - 1 + 1 = 0 + 1
or x = 1 is the required solution.
Check Put x = 1 in L.H.S.
∴ LHS = x - 1 = 1 - 1 = 0 = RHS

(b) x + 1 = 0
Answer:
Given equation is x + 1 = 0
Subtracting 4 from both sides,
x + 1 - 1 = 0 - 1
or x = - 1 is the required solution.
Check Put x = - 1 in L.H.S.
∴ L.H.S. = x + 1 = - 1 + 1 = 0 = R.H.S.

(c) x - 1 = 5
Answer:
x - 1 = 5
Adding 1 to both sides,
x - 1 + 1 = 5 + 1
or x = 6, is the required solution.
Check Put x = 6 in L.H.S..
L.H.S. = 6 - 1 = 5 = R.H.S.

(d) x + 6 = 2
Answer:
x + 6 = 2
Subtracting 6 from both sides,
x + 6 - 6 = 2 - 6
or x = - 4 is the required solution.
Check Put x = - 4 in L.H.S.
∴ L.H.S. = x + 6 = - 4 + 6
= 2 = R.H.S.

(e) y - 4 = - 7
Answer:
Given equation is y - 4 = - 7
Adding 4 to both sides,
or y = - 3 is the required solution.
Check Put y = - 3 in L.H.S.
L.H.S. = y - 4 = - 3 - 4 = - 7 = R.H.S.

(f) y - 4 = 4
Answer:
Given equation is y - 4 = 4
Adding 4 to both sides,
y - 4 + 4 = 4 + 4
or y = 8 is the required solution.
Check Put y = 0 in L.H.S.
∴ L.H.S. = y + 4 = 0 + 4 = 4 = R.H.S.

(g) y + 4 = 4
Answer:
Given equation is y + 4 = 4
Subtracting 4 from both sides,
y + 4 - 4 = 4 - 4
y = 0 is the required solUtion.
Check Put y = 0 in L.H.S.
∴ L.H.S. = y + 4 = 0 + 4 = 4 = R.H.S.

(h) y + 4 = - 4
Answer:
Given equation is y + 4 = - 4
Subtracting 4 from both sides,
y + 4 - 4 = - 4 - 4
y = - 8 is the required solution.
Check Put y = - 8 in L.H.S.
L.HS. = - 8 + 4 = - 4 = R.H.S.

Question 2.
Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42
Answer:
Given equation is 3l = 42
Dividing both sides by 3,
\(\frac{3 l}{3}=\frac{42}{3}\)
⇒ l = 14 is the required solution.
Check Put 1 = 14 in L.H.S.
∴ L.H.S. = 3l = 3 × 14 = 42 = R.H.S.

(b) \(\frac{b}{2}\) = 6
Answer:
Given equation is y = 6
Multiplying both sides by 2,
2 × \(\left(\frac{b}{2}\right)\) = 2 × 6
⇒ b = 12 is the required solution.
Check Put b = 12 in L.H.S.
∴ L.H.S. = \(\frac{b}{2}=\frac{12}{2}\) = 6 = R.H.S.

(c) \(\frac{p}{7}\) = 4
Answer:
Given equation is y = 4
Multiplying both sides by 7,
7 × \(\left(\frac{p}{7}\right)\) = 7 × 4
⇒ p = 28 is the required solution.
Check Put p = 28 in L.H.S.
L.H.S. = \(\frac{p}{7}\) = \(\frac{28}{7}\) = 4 = R.H.S.

(d) 4x = 25
Answer:
Given equation is 4x = 25
Dividing both sides by 4,
\(\frac{4 x}{4}=\frac{25}{4}\)
⇒ x = \(\frac{25}{4}\) is the required solution.
Check Putting x = \(\frac{25}{4}\) in L.H.S.
L.H.S. = 4x = 4 × \(\frac{25}{4}\) = 25 = R.H.S.

(e) 8y = 36
Answer:
Given equation is 8y = 36
Dividing both sides by 8,
\(\frac{8 y}{8}=\frac{36}{8}\)
⇒ y = \(\frac{36}{8}\) = \(\frac{36 \div 4}{8 \div 4}\)
= \(\frac{9}{2}\) is the required solution.
Check Put y = \(\frac{9}{2}\) in L.H.S.
L.H.S. = 8y = 8 × \(\frac{9}{2}\) = \(\frac{72}{2}\)
= 36 = R.H.S.

(f) \(\frac{z}{3}=\frac{5}{4}\)
Answer:
Given equation is \(\frac{\mathbf{z}}{3} = \frac{5}{4}\)
Multiplying both sides by 3,
3 × \(\left(\frac{z}{3}\right)\) = 3 × \(\frac{5}{4}\)
⇒ z = \(\frac{15}{4}\) is the required solution.
Check Put z = \(\frac{15}{4}\) in L.H.S.
L.H.S. = \(\frac{z}{3}=\frac{15}{4} \times \frac{1}{3}=\frac{5}{4}\) = R.H.S.

(g) \(\frac{a}{5}=\frac{7}{15}\)
Answer:
Given equation is \(\frac{a}{5}=\frac{7}{15}\)
Multiplying both sides by 5,
5 × \(\frac{a}{5}\) = 5 × \(\frac{7}{15}\)
⇒ a = \(\frac{35}{15}=\frac{35 \div 5}{15 \div 5}=\frac{7}{3}\)
is the required solution.
Put a = \(\frac{7}{3}\) in L.H.S.
L.H.S. = \(\frac{a}{5}=\frac{7}{3} \times \frac{1}{5}=\frac{7}{15}\) = R.H.S.

(h) 20t = - 10
Answer:
Given equation is 20t = - 10
Dividing both sides by 20,
\(\frac{20 t}{20}=-\frac{10}{20}\)
⇒ t = -\(\frac{1}{2}\) is the required solution.
Check Put t = -\(\frac{1}{2}\) in L.H.S.
L.H.S. = 20t = 20 × \(\left(-\frac{1}{2}\right)\) = - 10 = R.H.S.

Question 3.
Give the steps you will use to separate the variable and then solve the equation:
(a) 3n - 2 = 46
Answer:
Given equation is 3n - 2 = 46
Add 2 to both sides,
3n - 2 + 2 = 46 + 2
or 3n = 48
Divide both sides by 3,
or \(\frac{3 n}{3}=\frac{48}{3}\) = 48
⇒ n = 16 is the required solution.
Check Put n = 16 in L.H.S.
L.H.S. = 3n - 2 = 3 × 16 - 2
= 48 - 2 = 46 = R.H.S.

(b) 5m + 7 = 17
Answer:
Given equation is 5m + 7 = 17
Substract 7 from both sides,
5m + 7 - 7 = 17 - 7 ⇒ 5m = 10
Divide both sides by 5,
\(\frac{5 m}{5}=\frac{10}{5}\)
⇒ m = 2 is the required solution.
Check Put m = 2 in L.H.S.
L.H.S. = 5m + 7 = 5 × 2 + 7
= 17 = R.H.S.

(c) \(\frac{20 p}{3}\) = 40
Answer:
Given equation is \(\frac{20 p}{3}\) = 40
Multiply both sides by 3,
3 × \(\frac{20 p}{3}\) = 3 × 40 ⇒ 20p = 120
Divide both sides by 20,
\(\frac{20 p}{20}=\frac{120}{20}\)
⇒ p = 6 is the required solution.
Check Put p = 6 in L.H.S.
L.H.S. = \(\frac{20 p}{3}=\frac{20 \times 6}{3}=\frac{120}{3}\)
= 40 = R.H.S.

(d) \(\frac{3 p}{10}\) = 6
Answer:
Given equation is \(\frac{3 p}{10}\) = 6
Multiply both sides by 10,
10 × \(\frac{3 p}{10}\) = 10 × 6 ⇒ 3p = 60
Divide both sides by 3,
\(\frac{3 p}{3}\) = \(\frac{60}{3}\) = 20
is the required solution.
Check Put p = 20 in L.H.S.
L.H.S. = \(\frac{3 p}{10}=\frac{3 \times 20}{10}=\frac{60}{10}\)
= 6 = R.H.S.

Question 4.
Solve the following equations :
(a) 10p = 100
Answer:
Given equation is 10p = 100
Dividing both sides by 10,
\(\frac{10 p}{10}=\frac{100}{10}\)
⇒ p = 10 is the required solution.
Check Put p = 10 in L.H.S.
L.H.S. = 10p = 10 × 10 = 100
= R.H.S.

(b) 10p + 10 = 100
Answer:
Given equation is 10p + 10 = 100
Subtract 10 from both sides,
10p + 10 - 10 = 100 - 10
⇒ 10p = 90
Divide both sides by 10,
\(\frac{10 p}{10}=\frac{90}{10}\) ⇒ p
= 9 is the required solution.
Check Put p = 9 in L.H.S.
L.H.S. = 10p + 10 = 10 × 9 + 10 = 90 + 10 = 100 = R.H.S.

(c) \(\frac{p}{4}\) = 5
Answer:
Given equation is \(\frac{p}{4}\) = 5
Multiply both sides by 4,
4 × \(\frac{p}{4}\) = 4 × 5
⇒ p = 20 is the required solution.
Check Put p = 20 in L.H.S.
L.H.S. = \(\frac{p}{4}\) = \(\frac{20}{4}\) = 5 = R.H.S.

(d) \(\frac{-p}{3}\) = 5
Answer:
Given equation is = 5
Multiply both sides by - 3,
(- 3) × \(\left(\frac{-p}{3}\right)\) = (-3) × 5
⇒ p = - 15 is the required solution.
Check Put p = - 15 in L.H.S.
L.H.S. = \(\frac{-p}{3}=-\left(\frac{-15}{3}\right)=\frac{15}{3}\)
= 5 = R.H.S.

(e) \(\frac{3 p}{4}\) = 6
Answer:
Given equation is \(\frac{3 p}{4}\) = 6
Multiply both sides by 4,
4 × \(\frac{3 p}{4}\) = 4 × 6 ⇒ 3p = 24
Divide both sides by 3,
\(\frac{3 p}{3}\) = \(\frac{24}{3}\)
⇒ p = 8 is the required solution.
Check Put p = 8 in L.H.S.
L.H.S. = \(\frac{3 p}{4}=\frac{3 \times 8}{4}=\frac{24}{4}\)
= 6 = R.H.S.

(f) \(\frac{5}{2}\)x = \(\frac{25}{4}\)
Answer:
Given equation is 3s = - 9
Divide both sides by 3,
\(\frac{3 s}{3}=\frac{0}{3}\)
⇒ s = - 3 is the required solution.
Check Put s = - 3 in L.H.S.
L.H.S. = 3s = 3 × (- 3) = - 9 = R.H.S.

(g) 7m + \(\frac{19}{2}\) = 13
Answer:
Given equation is 3s + 12 = 0
Subtract 12 from both sides,.
3s + 12 - 12 = 0 - 12
⇒ 3s = - 12
Divide both sides by 3,
\(\frac{3 s}{3}=\frac{-12}{3}\)
⇒ s = - 4 is the required solution.
Check Put s = - 4 in L.H.S.
L.H.S. = 3s + 12 = 3 × (- 4) + 12
= - 12 + 12 = 0 = R.H.S.

(h) 3s = 0
Answer:
Given equation is 3s = 0
Divide both sides by 3,
\(\frac{3 s}{3}=\frac{0}{3}\)
⇒ s = 0 is the required solution.
Check Put s = 0 in L.H.S.
L.H.S. = 3s = 3 × 0 = 0 = R.H.S.

(i) 2q = 6 = 0
Answer:
Given equation is 2q = 6
Divide both sides by 2,
\(\frac{2 q}{2}=\frac{6}{2}\)
⇒ q = 3 is the required solution.
Check Put q = 3 in L.H.S.
L.H.S. = 2q = 2 × 3 = 6 = R.H.S.

(j) 2q - 6 = 0
Answer:
Given equation is 2q - 6 = 0
Add 6 to both sides,
2q,-6 + 6 = 0 + 6 ⇒ 2q = 6
Divide both sides by 2,
\(\frac{2 q}{2}=\frac{6}{2}\)
⇒ q = 3 is the required solution.
Check Put q = 3 in L.H.S.
L.H.S. = 2q - 6 = 2 × 3 - 6
= 6 - 6 = 0 = R.H.S.

(k) 2q + 6 = 0
Answer:
Given equation is 2q + 6 = 0
Subtract 6 from both sides,
2q + 6 - 6 = 0 - 6 ⇒ 2q = - 6
Divide both sides by 2,
\(\frac{2 q}{2}=-\frac{6}{2}\)
⇒ q = - 3 is the required solution.
Check Put q = - 3 in L.H.S.
L.H.S. = 2q + 6 = 2 × (- 3) + 6
= -6 + 6 = 0 = R.H.S.

(l) 2q + 6 = 12
Answer:
Given equation is 2q + 6 = 12
Subtract 6 from both sides,
2q + 6 - 6 = 12 - 6 ⇒ 2q = 6
Divide both sides by 2,
\(\frac{2 q}{2}=\frac{6}{2}\)
⇒ q - 3 is the required solution.
Check Put q = 3 in L.H.S.
L.H.S. = 2q + 6 = 2 × 3 + 6
= 6 + 6 = 12 = R.H.S.