SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.3 Textbook Exercise Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.3

Question 1.

Solve the following equations and check the result.

(i) 5x – 17 = 18

Answer:

Given 5x – 17 = 18

⇒ 5x – 17 + 17 = 18 + 17 (Add 17 on both sides)

⇒ 5x = 35

⇒ \(\frac{5 x}{5}\) = \(\frac{35}{5}\) (Divide by 5 on both sides)

⇒ x = 7

Check:

Substitute x = 7 in 5x – 17 = 18

LHS = 5x – 17

= 5(7) – 17

= 35 – 17 = 18 = RHS

Hence verified.

(ii) 29 – 7y = 1

Answer:

Given 29 – 7y = 1

⇒ 29 – 7y – 29 = 1 – 29 (Subtract 29 on both sides)

⇒ – 7y = -28 .

⇒ \(\frac{-7 y}{-7}\) = \(\frac{-28}{-7}\) (Divide by – 7 on both sides)

⇒ y = 4

Check:

Substitute y = 4 in

29 – 7y = 1

LHS = 29 – 7y

= 29 – 7(4)

= 29 – 28 = 1 = RHS

Hence verified.

(iii) a – 2.3 = 1.5

Answer:

Given a – 2.3 = 1.5

⇒ a – 2.3 + 2.3 = 1.5 + 2.3 (Add 2.3 ort both sides)

⇒ a = 3.8

Check:

Substitute a = 3.8 in a – 2.3 =1.5

LHS = a – 2.3

= 3.8 – 2.3

= 1.5 = RHS

Hence verified.

(iv) b + 3\(\frac{1}{2}\) = \(\frac{7}{4}\)

Answer:

(v) \(\frac{7 p}{10}\) + 9 = 15

Answer:

(vi) 6(q – 5) = 42.

Answer:

Given 6(q – 5) = 42

⇒ \(\frac{6(q-5)}{6}\) = \(\frac{42}{6}\) (Divide by 6 on both sides)

⇒ q – 5 = 7

⇒ q – 5 + 5 = 7 + 5 (Add 5 on both sides)

⇒ q = 12

Check:

Substitute q = 12 in

6(q – 5) = 42

LHS = 6(q – 5)

= 6(12 – 5)

= 6(7) = 42 = RHS

Hence verified.

(vii) – 3(m + 5) + 1 = 13

Answer:

Given – 3(m + 5)4-1 = 13

⇒ – 3(m + 5)+ 1 – 1 = 13 – 1 (Subtract 1 on both sides)

⇒ – 3(m + 5) = 12

⇒ \(\frac{-3(m+5)}{-3}\) = \(\frac{12}{-3}\) (Divide by – 3 on both sides)

⇒ m + 5 = – 4

⇒ m + 5 – 5 = – 4 – 5 (Subtract 5 on both sides)

⇒ m = – 9

Check:

Substitute m = – 9 in – 3(m + 5) + 1 = 13

LHS = – 3(m + 5) + 1

= – 3(- 9 + 5) + 1 = (- 3 × – 4) + 1

= + 12 + 1 = 13 = RHS

Hence verified.

(viii) \(\frac{n}{2}+\frac{n}{3}+\frac{n}{5}\) = 31

Answer:

Question 2.

Solve the following equations and check the result.

(i) 3(p – 7) – 4 = 5

Answer:

Given 3(p – 7) – 4 = 5

⇒ 3(p – 7) – 4 + 4 = 5 + 4 (Add 4 on both sides)

⇒ 3(p – 7) = 9

⇒ \(\frac{3(p-7)}{3}\) = \(\frac{9}{3}\) (Divide by 3 on both sides)

⇒ p – 7 = 3

⇒ p – 7 + 7 = 3 + 7 (Add 7 on both sides)

⇒ p = 10

Check:

Substitute p = 10 in

3(p – 7) – 4 = 5

LHS = 3(p – 7) – 4.

= 3(10 – 7) – 4.

= 3 × 3 – 4

= 9 – 4 = 5 = RHS

Hence verified.

(ii) 5(q – 3) – 3(q – 2) = 0

Answer:

Given 5(q – 3) – 3(q – 2) = 0

⇒ 5q – 15 – 3q + 6 = 0 (Distributive property)

⇒ 2q – 9 = 0 .

⇒ 2q – 9 + 9 = 0 + 9 (Add 9 on both sides)

⇒ 2q = 9

⇒ \(\frac{2 \mathrm{q}}{2}\) = \(\frac{9}{2}\) (Divide by 2 on both sides)

⇒ q = \(\frac{9}{2}\)

Check : Substitute q = \(\frac{9}{2}\) in

5(q – 3) – 3(q – 2) = 0

LHS = 5(q – 3) – 3(q – 2)

= 5\(\left(\frac{9}{2}-3\right)\) – 3\(\left(\frac{9}{2}-2\right)\)

= 5\(\left(\frac{9-6}{2}\right)\) – 3\(\left(\frac{9-4}{2}\right)\)

= 5 × \(\frac{3}{2}\) – 3 × \(\frac{5}{2}\)

= \(\frac{15}{2}-\frac{15}{2}\) = 0 = RHS

Hence verified.

(iii) 4x – 0.3x – 1.2 = 0.6

Answer:

Given 0.4x – 0.3x – 1.2 = 0.6

⇒ 0.1 x – 1.2 = 0.6

⇒ 0.1 x – 1.2 + 1.2 = 0.6 + 1.2 (Add 1.2 on both sides)

⇒ 0.1 x = 1.8

⇒ \(\frac{1 \mathrm{x}}{10}\) = \(\frac{18}{10}\)

(Multiply by 10 on both sides)

⇒ x = 18

Check:

Substitute x = 18 in

0.4x – 0.3x – 1.2 = 0.6

LHS = 0.4x – 0.3x – 1.2

= 0.4(18) – 0.3(18) – 1.2

= 7.2 – 5.4 – 1.2

= 7.2 – 6.6 = 0.6 = RHS

Hence verified.

(iv) 4(3y + 4) = 7.6

Answer:

Given 4(3y + 4) = 7.6.

⇒ \(\frac{4(3 y+4)}{4}\) = \(\frac{7.6}{4}\) (Divide by 4 on both sides)

⇒ 3y + 4 = 1.9

⇒3y + 4 – 4= 1.9 – 4 (Subtract 4 on both sides)

⇒ 3y = – 2.1

⇒ \(\frac{3 y}{3}\) = – \(\frac{2.1}{3}\) (Divide by 3 on both sides)

⇒ y = – 0.7

Check:

Substitute y = – 0.7 in

4(3y + 4) = 7.6

LHS = 4(3y + 4)

= 4[3 (- 0.7) + 4]

= 4[- 2.1 + 4]

= 4 × 1.9 = 7.6 = RHS

Hence verified.

(v) 20 – (2r – 5) = 25

Answer:

Given 20 – (2r – 5) = 25

(Subtract 20 on both sides)

⇒ – (2r – 5) = 5

⇒ – (2r -5) × – 1 = 5 × – 1

(Multiply by – 1 on both sides)

⇒ 2r – 5 = 5

⇒ 2r – 5 + 5= – 5 + 5 (Add 5 on both sides)

⇒ 2r = 0

⇒ \(\frac{2 \mathrm{r}}{2}\) = \(\frac{0}{2}\) (Divide by 2 on both sides)

⇒ r = 0

Check:

Substitute r = 0 in

20 – (2r – 5) = 25

LHS = 20 – (2r – 5)

= 20 – [2(0) – 5]

= 20 – [0 – 5]

= 20 – (- 5)

= 20 + 5 = 25 = RHS

Hence verified.

(vi) 3(5 – t) – 2(t – 2) = – 1

Answer:

Given 3(5 – t) – 2(t – 2) = – 1

⇒ 15 – 3t – 2t + 4 .= – 1 (Distributive property)

⇒ 19 – 5t = – 1

(Subtract 19 on both sides)

⇒ – 5t = – 20

⇒ \(\frac{-5 t}{-5}\) = \(\frac{-20}{-5}\) (Divide by – 5 on both sides)

⇒ t = 4

Check:

Substitute t = 4 in

3(5 -1) – 2(t – 2) = – 1

LHS = 3(5 – t) – 2(t – 2)

= 3(5 – 4) – 2(4 – 2)

= 3(1) – 2(2)

= 3 – 4 = – 1 = RHS

Hence verified.

(vii) 3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16

Answer:

Given 3(2k + 1) – 2(k T 5) – 5(5 – 2k) = 16

⇒ 6k + 3-2k + 10-25 + 10k = 16 (Distributive property)

⇒ 14k – 12 = 16

⇒ 14k – 12 + 12 – 16 + 12 (Add 12 on both sides)

⇒ 14k = 28

⇒ \(\frac{14 \mathrm{k}}{14}\) = \(\frac{28}{14}\) (Divide by 14 on both sides)

⇒ k = 2

Check:

Substitute k = 2 in

3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16

LHS = 3(2k + 1) – 2(k – 5) – 5(5 – 2k)

= 3[2 × 2 + 1] – 2[2 – 5] – 5[5 – 2 × 2]

= 3[4 + 1] – 2(- 3) – 5(5 – 4)

= 15 + 6 – 5 – 16 = RHS

Hence verified.

(viii) \(\frac{3 \mathrm{~m}}{4}\) – 5m – \(\frac{3}{4}\)

Answer:

⇒ – 17m – 3 = 48

⇒ – 17m – 3 + 3 =48 + 3 (Add 3 on both sides)

⇒ – 17m = 51

⇒ \(\frac{-17 \mathrm{~m}}{-17}\) = \(\frac{51}{-17}\) (Divide by – 17 on both sides)

⇒ m = – 3

(ix) \(\frac{4 n}{5}+\frac{n}{4}-\frac{n}{2}=\frac{11}{10}\)

Answer:

(x) \(\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3 x}{10}=\frac{1}{5}\)

Answer:

Question 3.

Write any three equivalent equations having the solution x = 2.

Answer:

Question 4.

Write any three equivalent equations having the solution a = – 5.

Answer: