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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.1

Question 1.

Fill in the blanks

Question (i)

The value of x in the equation x +5 12 ¡s ……….

Answer:

7

Hint:

Given,

x + 5 = 12

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

Value of x is 7

Question (ii)

The value ofy 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)

Question (iii)

The value of m in the equation 8m = 56 is ………

Answer:

7

Hint:

Given,

8m = 56

Divided by 8 on both sides

\(\frac{8xm}{8}\) = \(\frac{56}{8}\)

∴ m = 7

Question (iv)

The value ofp in the equation \(\frac{2p}{3}\) = 10 is ……….

Answer:

1

Hint:

Given,

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

Multiplying by 3 on both sides,

∴ p = 15

Question (v)

The linear equation in one variable has ……… Solution.

Answer:

one.

Question 2.

Say True or False.

Question (i)

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

Answer:

True

Question (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)

Hint:

a. \(\frac{x}{2}\) = 10,

multiplying by 2 on both sides, we get

\(\frac{x}{2}\) x 2 = 10 x 2 ⇒ x = 20

b. 20 = 6x – 4

by transposition ⇒ 20 + 4 = 6x

6x = 24

dividing by 6 on both sides,

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

∴ 3x = 8

∴ x = \(\frac{8}{3}\)

d. 7x – 4 – 8x = 20

by transposing – 4 to other side,

7x – 8x = 20 + 4

– x = 24

∴ x = – 24

\(\frac{4}{11}\) – 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:

Question (i)

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

Solution:

Transposing -4 to other side, it becomes +4

∴ \(\frac{2x}{3}\) = \(\frac{10}{3}\) + 4

Taking LCM & adding,

\(\frac{2x}{3}\) = \(\frac{10}{3}\) + \(\frac{4}{1}\) = \(\frac{10+12}{3}\) = \(\frac{22}{3}\)

\(\frac{2x}{3}\) = \(\frac{22}{3}\)

Multiplying by 3 on both sides

⇒ 2x = 22

dividing by 2 on both sides,

We get \(\frac{2x}{2}\) = \(\frac{22}{2}\)

∴ x = 11

Question (ii)

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

Solution:

Transposing \(\frac{1}{6}\) to the other side,

y – 3y = \(\frac{2}{3}\) – \(\frac{1}{6}\)

Taking LCM,

– 2y = \(\frac{2}{3}\) – \(\frac{1}{6}\) = \(\frac{2×2-1}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

∴ – 2y = \(\frac{1}{2}\) ⇒ 2y = – \(\frac{1}{2}\)

dividing by 2 or both sides.

Question (iii)

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

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

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

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

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

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:

Question (i)

-3(4x + 9) = 21

Solution:

Expanding the bracket,

-3 × 4x + (-3) × 9 = 21

-12x + (-27) = 21

-12x – 27 = 21

Transposing – 27 to other side, it becomes +27

-12x = 21 + 27 = 48

12x = 48 ⇒ 12x = -48

Dividing by 12 on both sides

⇒ x = – 4

Question (ii)

20 – 2 ( 5 – p) = 8

Solution:

Expanding the bracket,

20 – 2 x 5 – 2 x (-p) = 8

20 – 10 + 2 + p = 8 (-2 x -P = 2p)

10 + 2p = 8 transporting 10 to other side

2P = 8 – 10 = -2

∴ 2p = -2

∴ p = -1

Question (iii)

(7x – 5) – 4(2 + 5x) = 10(2 – x)

Solution:

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:

Question (i)

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

Solution:

\(\frac{3x-2}{4}\) – \(\frac{(x-3)}{5}\)

Taking LCM on LHS, [LCM of 4 & 5 is 20]

∴ 11x + 2 = -20

∴ 11x = – 20 – 2 = – 22

x = \(\frac{-22}{11}\) = -2

x = -2

Question (ii)

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

Solution:

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

Cross multiplying, we get

∴ (m + 9) x 3 = 5 x (3m + 15)

m x 3 + 9 x 3 = 5 x 3m + 5 x 15

Transporting 3m & 75, we get

27 – 75 = 15m – 3m

-48 = 12m

⇒ m = -4