Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.6 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.6

Question 1.

Fill in the blanks:

(i) The value of x in the equation x + 5 = 12 is ________ .

Answer:

7

Hint:

Given, x + 5 = 12

x = 12 – 5 = 7 (by transposition method)

Value of x is 7

(ii) The value of y in the equation y – 9 = (-5) + 7 is ________ .

Answer:

11

Hint:

Given, y – 9 = (-5) + 7

y – 9 = 7 – 5 (re-arranging)

y – 9 = 2

∴ y = 2 + 9 = 11 (by transposition method)

(iii) The value of m in the equation 8m = 56 is ________ .

Answer:

7

Hint:

Given, 8m = 56

Divided by 8 on both sides

\(\frac{8 \times m}{8}=\frac{56}{8}\)

∴ m = 7

(iv) The value of p in the equation \(\frac{2 p}{3}\) = 10 is ________ .

Answer:

15

Hint:

Given, \(\frac{2 p}{3}\) = 10

Multiplying by 3 on both sides

Dividing by 2 on both sides

∴ p = 15

(v) The linear equation in one variable has ________ solution.

Answer:

one

Question 2.

Say True or False.

(i) The shifting of a number from one side of an equation to other is called transposition.

Answer:

True

(ii) Linear equation in one variable has only one variable with power 2.

Answer:

False

[Linear equation in one variable has only one variable with power one – correct statement]

Question 3.

Match the following

(A) (i),(ii), (iv) ,(iii),(v)

(B) (iii), (iv), (i) ,(ii), (v)

(C) (iii),(i) ,(iv), (v), (ii)

(D) (iii) , (i) , (v) ,(iv) ,(ii)

Answer:

(C) (iii),(i) ,(iv), (v), (ii)

a. \(\frac{x}{2}\) = 10, multiplying by 2 on both sides, we get

\(\frac{x}{2}\) × 2 = 10 × 2 ⇒ x = 20

b. 20 = 6x – 4 by transposition ⇒ 20 + 4 =6x

6x = 24 dividing by 6 on both sides,

\(\frac{6 x}{6}=\frac{24}{6}\) ⇒ x = 4

c. 2x – 5 = 3 – x

By transposing the variable ‘x’, we get

2x – 5 + x = 3

by transposing – 5 to other side,

2x + x = 3 + 5

∴ x = \(\frac{8}{3}\)

d. 7x – 4 – 8x = 20

by transposing – 4 to other side,

7x – 8x = 20 + 4

– x = 24

∴ x = – 24

e. \(\frac{4}{11}-x=\frac{-7}{11}\)

Transposing \(\frac{4}{11}\) to other side,

– x = \(\frac{-7}{11} \frac{-4}{11}=\frac{-7-4}{11}=\frac{-11}{11}\) = – 1

∴ – x = – 1 ⇒ x = 1

Question 4.

Find x:

(i) \(\frac{2 x}{3}-4=\frac{10}{3}\)

Answer:

Transposing – 4 to other side, it becomes + 4

(ii) \(y+\frac{1}{6}-3 y=\frac{2}{3}\)

Answer:

Transposing \(\) to the other side,

(iii) \(\frac{1}{3}-\frac{x}{3}=\frac{7 x}{12}+\frac{5}{4}\)

Answer:

Transposing \(\frac{-x}{3}\) to the other side, it becomes + \(\frac{x}{3}\)

∴ \(\frac{1}{3}=\frac{7 x}{12}+\frac{5}{4}+\frac{x}{3}\)

Transposing \(\frac{5}{4}\) to the other side, it becomes \(\frac{-5}{4}\)

\(\frac{1}{3}-\frac{5}{4}=\frac{7 x}{12}+\frac{x}{3}\)

Multiply by 12 throughout [we look at the denominators 3, 4, 12, 3 and take the LCM, which is 12]

4 – 15 = 7x + x × 4

– 11 = 7x + 4x

11x = – 11

x = -1

Question 5.

Find x

(i) -3(4x + 9) = 21

Answer:

Expanding the bracket,

-3 × 4 + (-3) × 9 = 21

∴ -12x + (-27) = 21

– 12x – 27 = 21

Transposing – 27 to other side, it becomes +27

– 12 x = 21 + 27 = 48

∴ – 12x = 48 ⇒ 12x = – 48

Dividing by 12 on both sides

(ii) 20 – 2 (5 – p) = 8

Answer:

Expanding the bracket,

20 – 2 × 5 – 2 × (-p) = 8

20 – 10 + 2p = 8

(- 2 × – p = 2p)

10 + 2p = 8 transposing lo to other side

2p = 8 – 10 = – 2

∴ 2p = – 2

∴ p = – 1

(iii) (7x – 5) – 4(2 + 5x) = 10(2 – x)

Answer:

Expanding the brackets,

7x – 5 – 4 × 2 – 4 × 5x = 10 × 2 + 10 × (-x)

7x – 5 – 8 – 20x = 20 – 10x

7x – 13 – 20x = 20 – 10x

Transposing 10x & – 13, we get

7x – 13 – 20x + 10x = 20

7x – 20x + 10x = 20 + 13, Simplifying,

– 3x = 33

3x = – 33

x = \(\frac{-33}{3}\) = – 11

x = – 11

Question 6.

Find x and m:

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

Answer:

(ii) \(\frac{m+9}{3 m+15}=\frac{5}{3}\)

Answer:

Cross multiplying, we get

∴ (m + 9) × 3 = 5 × (3m + 15)

m × 3 + 9 × 3 = 5 × 3m + 5 × 15

27 – 75 = 15m – 3m

– 48 = 12m

⇒ m = – 4