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

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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 ……….
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 ……….
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 ………
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 ……….
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.
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.
True

Question (ii)
Linear equation in one variable has only one variable with power 2.
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)
(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
$$\frac{2x}{3}$$ = $$\frac{10}{3}$$ + $$\frac{4}{1}$$ = $$\frac{10+12}{3}$$ = $$\frac{22}{3}$$
$$\frac{2x}{3}$$ = $$\frac{22}{3}$$
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