Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.1
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.1
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
8xm/8 = 56/8
∴ m = 7
Question (iv)
The value ofp in the equation 2p/3 = 10 is ……….
Answer:
1
Hint:
Given,
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. x/2 = 10,
multiplying by 2 on both sides, we get
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,
6x/6 = 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 = 8/3
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 = −33/3 = -11
x = -11
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