NCERT Solutions Class 8 Mathematics
Chapter – 9 (Algebraic Expressions and Identities)
The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 9 Algebraic Expressions and Identities Exercise 9.1 has been provided here to help the students in solving the questions from this exercise.
Chapter 9: Algebraic Expressions and Identities
- NCERT Solution Class 8 Maths Ex – 9.2
- NCERT Solution Class 8 Maths Ex – 9.3
- NCERT Solution Class 8 Maths Ex – 9.4
- NCERT Solution Class 8 Maths Ex – 9.5
Exercise – 9.1
1. Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz2 – 3zy
(ii) 1 + x + x2
(iii) 4x2y2 – 4x2y2z2 + z2
(iv) 3 – pq + qr – p
(v) – xy
(vi) 0.3a – 0.6ab + 0.5b
Solution –
| Sl. No. | Expression | Term | Coefficient |
| (i) | 5xyz2 – 3zy | 5xyz2, -3zy | 5, -3 |
| (ii) | 1 + x + x2 | 1, x, x2 | 1, 1, 1 |
| (iii) | 4x2y2 – 4x2y2z2 + z2 | 4x2y2, -4 x2y2z2, z2 | 4, -4, 1 |
| (iv) | 3 – pq + qr – p | 3 -pq, qr -p | 3, -1, 1, -1 |
| (v) | |||
| (vi) | 0.3a – 0.6ab + 0.5b | 0.3a, -0.6ab, 0.5b | 0.3, -0.6, 0.5 |
2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x2 + x3 + x4 , 7 + y + 5x, 2y – 3y2 , 2y – 3y2 + 4y3 , 5x – 4y + 3xy, 4z – 15z2 , ab + bc + cd + da, pqr, p2q + pq2 , 2p + 2q
Solution – Let us first define the classifications of these 3 polynomials:
- Monomials contain only one term.
- Binomials contain only two terms.
- Trinomials contain only three terms.
| Binomial | Monomial | Trinomial | Polynomial |
| x + y | 1000 | 2y – 3y2 + 4y3 | x + x2 + x3 + x4 |
| 2y – 3y2 |
pqr | 5x – 4y + 3xy | ab + bc + cd + da |
| 4z – 15z2 | 7 + y + 5x | ||
| p2q + pq2 |
|||
| 2p + 2q |
3. Add the following.
(i) ab – bc, bc – ca, ca – ab
(ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
Solution –
(i) ab – bc, bc – ca, ca – ab
= (ab – bc) + (bc – ca) + (ca-ab)
= ab – bc + bc – ca + ca – ab
= ab – ab – bc + bc – ca + ca
= 0
(ii) a – b + ab, b – c + bc, c – a + ac
= (a – b + ab) + (b – c + bc) + (c – a + ac)
= a – b + ab + b – c + bc + c – a + ac
= a – a +b – b +c – c + ab + bc + ca
= 0 + 0 + 0 + ab + bc + ca
= ab + bc + ca
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
= 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
= (2p2q2 – 3pq + 4) + (5 + 7pq – 3p2q2)
= 2p2q2 – 3p2q2 – 3pq + 7pq + 4 + 5
= – p2q2 + 4pq + 9
(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
= (l2 + m2) + (m2 + n2) + (n2 + l2) + (2lm + 2mn + 2nl)
= l2 + l2 + m2 + m2 + n2 + n2 + 2lm + 2mn + 2nl
= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl
4.
(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q
Solution –
(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
= (12a – 9ab + 5b – 3) – (4a – 7ab + 3b + 12)
= 12a – 9ab + 5b – 3 – 4a + 7ab – 3b – 12
= 12a – 4a -9ab + 7ab +5b – 3b -3 -12
= 8a – 2ab + 2b – 15
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
= (5xy – 2yz – 2zx + 10xyz) – (3xy + 5yz – 7zx)
= 5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx
= 5xy – 3xy – 2yz – 5yz – 2zx + 7zx + 10xyz
= 2xy – 7yz + 5zx + 10xyz
(c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q
= (18 – 3p – 11q + 5pq – 2pq2 + 5p2q) – (4p2q – 3pq + 5pq2 – 8p + 7q – 10)
= 18 – 3p – 11q + 5pq – 2pq2 + 5p2q – 4p2q + 3pq – 5pq2 + 8p – 7q + 10
= 18 + 10 – 3p + 8p – 11q – 7q + 5pq+ 3pq- 2pq2 – 5pq2 + 5p2q – 4p2q
= 28 + 5p – 18q + 8pq – 7pq2 + p2q
