Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities InText Questions and Answers.
PSEB 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities InText Questions
Try These: [Textbook Page No. 139]
1. Write two terms which are like:
Question (i)
7xy
Solution:
14xy, 21xy
Question (ii)
4mn2
Solution:
8mn2 , – 11mn2
Question (iii)
21
Solution:
– 5l, 9l
Try These : [Textbook Page No. 142]
1. Can you think of two more such situations, where we may need to multiply algebraic expressions?
Solution:
1. Aarush purchased x notebooks and y pens. If cost of a notebook and a pen is same ₹ z, what amount has he to pay? → ₹ z (x + y)
2. Shailja wants to spread a carpet in her room having length (l + 5) m and breadth (b – 2) m. Find the area of the carpet. → (l + 5) (b – 2) m2
Try These: [Textbook Page No. 143]
1. Find 4x × 5y × 7z.
Solution:
4x × 5y × 7z = 4 × 5 × 7 × x × y × z
= 140 xyz
2. Find 4x × 5y × 7z. First find (4x × 5y) and multiply it by 7z; or first find (5y × 7z) and multiply it by 4x. Is the result the same? What do you observe? Does the order in which you carry out the multiplication matter?
Solution:
(4x × 5 y) = 4 × 5 × x × y
= 20xy
Now, 20xy × 7z = 20 × 7 × xy × z
= 140xyz … (i)
Also, (5y × 7z) = 5 × 7 × y × z = 35 yz
Now, 35yz × 4x = 35 × 4 × yz × x
= 140xyz … (ii)
Yes, the result is same.
We can conclude that product remains same if we change order of the terms.
3. Complete the table for area of a rectangle with given length and breadth.
Solution:
| length | breadth | area |
| 3x | 5y | 3x × 5y = 15xy |
| 9y | 4y2 | 9y × 4y2 = 36y3 |
| 4ab | 5bc | 4ab × 5be = 20ab2c |
| 2l2m | 3lm2 | 2l2m × 3lm2 = 6l3m3 |
Try These : [Textbook Page No. 144]
1. Find the product:
Question (i)
2x (3x + 5xy)
Solution:
= (2x × 3x) + (2x × 5xy)
= 6x2 + 10x2y
Question (ii)
a2 (2ab – 5c)
Solution:
= (a2 × 2ab) – (a2 × 5c)
= 2a3b – 5a2c
Try These: [Textbook Page No. 145]
1. Find the product: (4p2 + 5p + 7) × 3p
Solution:
(4p2 + 5p + 7) × 3p
= (4p2 × 3p) + (5p × 3p) + (7 × 3p)
= 12p3 + 15p2 + 21p
Try These : [Textbook Page No. 149]
1. Put -b in place of b in Identity (I). Do you get Identity (II)?
Solution:
Identity (I): (a + b)2 = a2 + 2ab + b2
Let us put (- b) instead of b [a + (- b)]2
= a2 + 2a (- b) + (- b)2
∴ (a – b)2
= a2 – 2ab + b2
Identity (II): (a – b)2 = a2 – 2ab + b2
Yes, we get Identity (II).
Try These : [Textbook Page No. 149]
1. Verify Identity (IV), for a = 2, b = 3, x = 5.
Solution:
Identity (IV):
(x + a) (x + b) = x2 + (a + b) x + ab
Substitute a = 2, b = 3 and x = 5
LHS
= (x + a) (x + b)
= (5 + 2) (5 + 3)
= (7)(8)
= 56
RHS
= x2 + (a + b) x + ab
= (5)2 + (2 + 3) × 5 + (2 × 3)
= 25 + (5) × 5 + (6)
= 25 + 25 + 6 = 56
∴ LHS = RHS
∴ The given identity is true for the given values.
2. Consider, the special case of Identity (IV) with a = b, what do you get ? Is it related to Identity (I)?
Solution:
When a = b (∴ Take y for both)
(x + a) (x + b) = x2 + (a + b) x + ab
Substitute a = y and b = y
(x + y)(x + y) = x2 + (y + y)x + (y × y)
= x2 + (2y) x + (y × y) = x2 + 2xy + y2
∴ Yes, it is the same as Identity ( I).
3. Consider, the special case of Identity (IV) with a = -c and b = -c. What do you get ? Is it related to Identity (II) ?
Solution:
Identity (IV):
(x + a)(x + b) = x2 + (a + b) x + ab
Substitute (- c) instead of a and (- c) instead of b,
(x – c) (x – c)
= x2 + [(-c) + (-c)]x + [(-c) × (-c)]
= x2 + [- 2c] x + (c2)
= x2 – 2cx + c2
∴ Yes, it is the same as Identity (II).
4. Consider the special case of Identity (IV) with b = – a. What do you get ? Is it related to Identity (III)?
Solution :
Identity (IV):
(x + a) (x + b) = x2 + (a + b) x + ab
Substitute (-a) instead of b,
(x + a) (x – a)
= x2 + [a + (- a)] x + [a × (- a)]
= x2 + (a – a) x + [- a2]
= x2 + (0) x – a2
= x2 – a2
∴ Yes, it is the same as Identity (III).