Algebraic Expressions and Identities Class 8 Extra Questions with Answers

These Class 8 Maths Extra Questions Chapter 9 Algebraic Expressions and Identities will help students prepare well for the exams.

Class 8 Maths Chapter 9 Extra Questions Algebraic Expressions and Identities

Class 8 Maths Algebraic Expressions and Identities Extra Questions

Question 1.
Simplify 2a(a – 2) – 2(a – 1).
Solution:
2a(a- 2) – 2(a – 1) = 2a² – 4a – 2a + 2 = 2a² – 6a + 2

Question 2.
What is the value of (x – 3y + 2z) + (2x + y – 4z)?
Solution:
(x – 3y + 2z) + (2x + y – 4z) = x + 2x + y – 3y + 2z – 4z = 3x – 2y – 2z.

Algebraic Expressions and Identities Class 8 Extra Questions

Question 1.
Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y, -3xz + 5x – 2xy.
Solution:
Writing the three expressions in separate rows, with like terms one below the other, we have
(Note xz is same as zx)
Thus, the sum of the expressions is 5xy + 9yz + 3zx + 5x – 4y. Note how the terms, – 4y in the second expression and 5x in the third expression, are carried over as they are, since they have no like terms in the other expressions.

Question 2.
Subtract 5x² – 4y² + 6y – 3 from 7x² – 4xy + 8y² + 5x – 3y.
Solution:

Question 3.
Complete the table for area of a rectangle with given length and breadth.

length	breadth	area
3x	5y	3x × 5y = 15xy
9y	4y2	………………………..
4ab	5bc	………………………..
2l2m	3lm2	………………………

Solution:

length	breadth	area
3x	5y	(3x) × (5y) = I5xy
9y	4y2	9y × 4y2 = 36y3
4ab	5bc	4ab × 5bc = 20 ab2c
2l2m	3lm2	2l2m × 3lm2 = 6l3m3

Question 4.
Find the volume of each rectangular box with given length, breadth and height.

length	breadth	height
(i) 2ax	3by	5cz
(ii) m2n	n2p	p2m
(iii) 2q	4q2	8q3

Solution:
Volume = length × breadth × height
Hence, for (i) volume = (2ax) × (3by) × (5cz)
= 2 × 3 × 5 × (ax) × (by) × (cz) = 30abcxyz
for (ii) volume = m²n × n²p × p²m
= (m² × m) × (n × n²) × (p × p²) = m³n³p³
for (iii) volume = 2q × 4q² × 8q³
= 2 × 4 × 8 × q × q² × q³ = 64q⁶

Question 5.
Simplify the expressions and evaluate them as directed :
i) x (x – 3) + 2 for x = 1,
i) 3y (2y – 7) – 3(y – 4) – 63 for y = -2
Solution:
i) x (x – 3) + 2 = x² – 3x + 2
For x = 1, x² – 3x + 2 = (1)² – 3 (1) + 2
= 1 – 3 + 2 = 3 – 3 = 0

ii) 3y(2y – 7) – 3(y – 4) – 6² = 6y² – 21y – 3y + 12 – 63
= 6y² – 24y – 51
For y = -2, 6y² – 24y – 51 = 6 (- 2)² – 24(- 2) – 51
= 6 × 4 + 24 × 2 – 51
= 24 + 48 – 51 = 72 – 51 = 21

Question 6.
Add.
i) 5m (3 – m) and 6m² – 13m
ii) 4y (3y² + 5y – 7) and 2 (y³ – 4y² + 5)
Solution:
i) First expression = 5m (3 – m) = (5m × 3) – (5m × m) =15m – 5m²
Now adding the second expression to it,15m – 5m² + 6m² – 13m = m² + 2m

ii) The first expression = 4y (3y² + 5y – 7) = (4y × 3y²) + (4y × 5y) + (4y × (-7))
= 12y³ + 20y² – 28y
The second expression = 2 (y³ – 4y² + 5) = 2y³ + 2 × (- 4y²) + 2 × 5
= 2y³ – 8y² + 10
Adding the two expressions,

Question 7.
Subtract 3pq (p – q) from 2pq (p + q).
Solution:

Question 8.
Multiply
i) (x – 4) and (2x + 3)
ii) (x – y) and (3x + 5y)
Solution:
i) (x – 4) × (2x + 3) = x × (2x + 3) – 4 × (2x + 3)
= (x × 2x) + (x × 3) – (4 × 2x) – (4 × 3)
= 2x² + 3x – 8x – 12 = 2x² – 5x – 12 (Adding like terms)

ii) (x – y) × (3x + 5y) = x × (3x + 5y) – y × (3x + 5y)
= (x × 3x) + (x × 5y)-(y × 3x)-(y × 5y)
= 3x² + 5xy – 3yx – 5y² = 3x² + 2xy – 5y² (Adding like terms)

Question 9.
Multiply
i) (a + 7) and (b – 5)
ii) (a² + 2b²) and (5a – 3b)
Solution:
i) (a + 7) × (b – 5) = a × (b – 5) + 7 × (b – 5)
= ab – 5a + 7b – 35
Note that there are no like terms involved in this multiplication.

ii) (a² + 2b²) × (5a – 3b) = a² (5a – 3b) + 2b² × (5a – 3b)
= 5a³ – 3a²b + 10ab² – 6b³
• Multiplying a binomial by a trinomial
(2a + 3) (3a² – 4a + 6)
each term in the first binomial has to multiply each term in trinomial, then (2a + 3) (3a² – 4a + 6) = 2a(3a² – 4a + 6) + 3(3a² – 4a + 6)
= 2a(3a)² + 2a(- 4a) + 2a(6) + 3(3a²) + 3(- 4a) + 3(6)
= 6a³ – 8a² + 12a + 9a² – 12a + 18 = 6a³ + a² + 18

Question 10.
Simplify (a + b) (2a – 3b + c) – (2a – 3b) c.
Solution:
We have
(a + b) (2a – 3b + c) = a (2a – 3b + c) + b (2a – 3b + c)
= 2a² – 3ab + ac + 2ab – 3b² + bc
= 2a² – ab – 3b² + be + ac (Note, -3ab and 2ab are like terms) and (2a – 3b) c = 2ac – 3bc
Therefore,
(a + b) (2a – 3b + c) – (2a – 3b) c = 2a² – ab – 3b² + be + ac – (2ac – 3bc)
= 2a² – ab – 3b² + be + ac – 2ac + 3bc
= 2a² – ab – 3b² + (be + 3bc) + (ac – 2ac)
= 2a² – 3b² – ab + 4bc – ac

Question 11.
Using the Identity (I), find
i) (2x + 3y)²
ii) 103²
Solution:
i) (2x + 3y)² = (2x)² + 2(2x) (3y) + (3y)² [Using the Identity (I)]
= 4x² + 12xy + 9y²
We may work out (2x + 3y)² directly.
(2x + 3y)² = (2x + 3y) (2x + 3y)
= (2x) (2x) + (2x) (3y) + (3y) (2x) + (3y) (3y)
= 4x² + 6xy + 6 yx + 9y² (as xy = yx)
= 4x² + 12xy + 9y²

ii) (103)² = (100 + 3)²
= 100² + 2 × 100 × 3 + 32 (Using Identity I)
= 10000 + 600 + 9 = 10609

Question 12.
Using Identity (II), find
i) (4p – 3q)²
ii) (4.9)²
Solution:
i) (4p – 3q)² = (4p)² – 2 (4p) (3q) + (3q)² [Using the Identity (II)]
= 16p² – 24pq + 9q²

ii) (4.9)² = (5.0 – 0.1)² = (5.0)² – 2 (5.0) (0.1) + (0.1)²
= 25.00 – 1.00 + 0.01 = 24.01

Question 13.
Using Identity (III), find
i) (\(\frac{3}{2}\)m + \(\frac{2}{3}\)n)(\(\frac{3}{2}\)m – \(\frac{2}{3}\)n)
ii) 983² – 17²
iii) 194 × 206
Solution:
i) (\(\frac{3}{2}\)m + \(\frac{2}{3}\)n)(\(\frac{3}{2}\)m – \(\frac{2}{3}\)n) = (\(\frac{3}{2}\)m)² – (\(\frac{2}{3}\)n)² = \(\frac{9}{4}\)m² – \(\frac{4}{9}\)mn²

ii) 983² – 17² = (983 + 17) (983 – 17) [Here a = 983, b – 17, a² – b² = (a + b) (a – b)]
Therefore, 983² – 17² = 1000 × 966 = 966000

iii) 194 × 206 =(200 – 6) × (200 + 6) = 200² – 6² = 40000 – 36 = 39964.

Question 14.
Use the Identity (x + a) (x + b) = x² + (a + b) x + ab to find the following:
i) 501 × 502
ii) 95 × 103
Solution:
i) 501 × 502 = (500 + 1) × (500 + 2) = 500² + (1 + 2) × 500 + 1 × 2
= 250000 + 1500 + 2 = 251502

ii) 95 × 103 = (100 – 5) × (100 + 3) = 100² + (-5 + 3) × 100 + (-5) × 3
= 10000 – 200 – 15 = 9785