SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.2 Textbook Exercise Questions and Answers.
AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.2
Question 1.
Give the steps you will use to separate the variables and then solve the equation.
(i) \(\frac{5 m}{3}\) = 10
Answer:
Given \(\frac{5 m}{3}\) = 10
⇒ \(\frac{5 m}{3}\) × 3 = 10 × 3 (Multiply with 3 on both sides)
⇒ 5m = 30
⇒ \(\frac{5 m}{5}\) = \(\frac{30}{5}\) (Divide both sides by 5)
⇒ m = 6
Check:
Substitute m = 6 in the given equation,
LHS = \(\frac{5 m}{3}\) = \(\frac{5(6)}{3}\) = \(\frac{30}{3}\) = 10 = RHS.
Hence verified.
(ii) 4M – 23 = 13
Answer:
Given 4n – 23 = 13
⇒ 4n = 36
⇒ \(\frac{4 n}{4}\) = \(\frac{36}{4}\) (Divide both sides by 4)
⇒ n = 9
Check:
Substitute n = 9 in the given equation.
LHS = 4n – 23
= 4(9) – 23
= 36 – 23 = 13 = RHS
Hence verified.
(iii) – 5 + 3x = 16
Answer:
Given – 5 + 3x = 16
⇒ – 5 + 3x + 5 = 16 + 5 (Add 5 on both sides)
⇒ 3x = 21
⇒ \(\frac{3 x}{3}\) = \(\frac{21}{3}\) (Divide by 3 on both sides)
⇒ x = 7
Check:
Substitute x = 7 in the given equation
LHS = – 5 + 3x
= – 5 + 3(7)
= – 5 + 21 = 16 = RHS
Hence verified.
(iv) 2(y – 1) =8
Answer:
Given 2(y – 1) = 8
⇒ 2y – 2 = 8 (Distributive property)
⇒ 2y – 2 + 2 = 8 + 2 (Add 2 on both sides)
⇒ 2y = 10
⇒ \(\frac{2 y}{2}\) = \(\frac{10}{2}\) (Divide by 2 on both sides)
⇒ y = 5
Check: Substitute y = 5 in the given equation
LHS = 2(y – 1)
= 2(5 – 1)
= 2 × 4 = 8 = RHS
Hence verified.
Question 2.
Solve the following simple equations and check the results.
(i) 3x = 18
Answer:
Given 3x = 18
⇒ \(\frac{3 x}{3}\) = \(\frac{18}{3}\) (Divide by 3 on both sides)
⇒ x = 6
Check: Substitute x = 6 in 3x = 18
LHS ⇒ 3x = 3(6) = 18 = RHS
Hence verified.
(ii) \(\frac{b}{7}\) = – 2
Sol.
Given \(\frac{b}{7}\) = – 2
⇒ \(\frac{b}{7}\) × 7 = – 2 × 7 (Multiply by 7 on both sides)
⇒ b = – 14
Check: Substitute b = – 14 in
\(\frac{b}{7}\) = – 2
LHS = \(\frac{b}{7}\) = \(\frac{- 14}{7}\) = – 2 = RHS
Hence verified
(iii) – 2x = – 10
Answer:
Given – 2x = – 10
⇒ \(\frac{-2 x}{-2}\) = \(\frac{-10}{-2}\) (Divide by – 2 on both sides)
⇒ x = 5
Check:
Substitute x = 5 in
– 2x = – 10
LHS = – 2x
= – 2(5) = – 10 = RHS
Hence verified.
(iv) 10 + 6a = 40
Answer:
Given 10 + 6a = 40
⇒ 10 + 6a – 10 = 40 – 10 (Subtract 10 on both sides)
⇒ 6a = 30
⇒ \(\frac{6a}{6}\) = \(\frac{30}{6}\) (Divide by 6 on both sides)
⇒ a = 5
Check: Substitute a = 5 in 10 + 6a = 40
LHS = 10 + 6a
= 10 + 6(5)
= 10 + 30
= 40 = RHS
Hence verified.
(v) – 7m = 21
Answer:
Given – 7m = 21
⇒ \(\frac{-7 m}{-7}\) = \(\frac{21}{-7}\) (Divide by – 7 on both sides)
⇒ m = – 3
Check:
Substitute m = in – 7m = 21
LHS = – 7m
= – 7 (- 3)
= 21 = RHS
Hence verified.
(iv) 4p + 7 = – 21
Answer:
Given 4p + 7 = – 21
⇒ 4p + 7 – 7 = – 21 – 7 (Subtract 7 on both sides)
⇒ 4p = – 28
⇒ \(\frac{4p}{4}\) = \(\frac{-28}{4}\) (Divide by 4 on both sides)
⇒ p = – 7
Check:
Substitute p = – 7 in 4p + 7 = – 21
LHS = 4p + 7
= 4(- 7) + 7
= – 28 + 7
= – 21 = RHS
Hence verified.
(vii) 3x – \(\frac{1}{3}\) = 5
Answer:
Check: Substitute x = \(\frac{16}{9}\) in 3x – \(\frac{1}{3}\) = 5
LHS = 3x – \(\frac{1}{3}\)
= 3\(\left(\frac{16}{9}\right)\) – \(\frac{1}{3}\)
= \(\frac{16}{3}-\frac{1}{3}\)
= \(\frac{16-1}{3}\) = \(\frac{15}{3}\)
= 5 = RHS
Hence verified.
(viii) 18 – 7n = – 3
Answer:
Given 18 – 7n = – 3
⇒ 18 – 7n – 18 = – 3 – 18 (Subtract 18 on both sides)
⇒ – 7n = – 21
⇒ \(\frac{-7 n}{-7 n}\) = \(\frac{-21}{-7}\) (Divide by – 7 on both sides)
⇒ n = 3
Check:
Substitute n = 3 in 18 – 7n = – 3
LHS = 18 – 7n
= 18 – 7(3)
= 18 – 21
= – 3 = RHS
Hence verified.
(ix) 3(k + 4) = 21
Sol.
Given 3(k + 4) = 21
⇒ 3k + 12 = 21 (Distributive property)
⇒ 3k + 12 – 12 = 21 – 12 (Subtract 12 on both sides)
⇒ 3k = 9
⇒ \(\frac{3 \mathrm{k}}{3}\) = \(\frac{9}{3}\) (Divide by 3 on both sides)
⇒ k = 3
Check:
Substitute k = 3 in 3(k + 4) = 21
LHS = 3(k + 4)
= 3(3 + 4)
= 3 × 7 = 21= RHS
Hence verified.
(x) 9 (a + 1) + 2 = 11
Answer:
Given 9(a + 1) + 2 = 11
⇒ 9(a + 1) + 2 – 2 = 11 – 2 (Subtract 2 on both sides)
⇒ 9(a + 1) = 9
⇒ \(\frac{9(a+1)}{9}\) = \(\frac{9}{9}\) (Divide by 9 on both sides)
⇒ a + 1 = 1
⇒ a + 1 – 1 = 1 – 1 (Subtract 1 on both sides)
⇒ a = 0
Check:
Substitute a = 0 in 9(a + 1) + 2 = 11
LHS = 9(a + 1) + 2
= 9(0 + 1) + 2
= 9(1) + 2
= 9 + 2 = 11 = RHS
Hence verified.