## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

**NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.1**

Ex 9.1 Class 8 Maths Question 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 – rp

(v) (frac { x }{ 2 }) + (frac { y }{ 2 }) – xy

(vi) 0.3a – 0.6ab + 0.5b

Solution:

Ex 9.1 Class 8 Maths Question 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:

Ex 9.1 Class 8 Maths Question 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) Given: ab – bc, bc – ca, ca – ab

We have

(ab – bc) + (bc – ca) + (ca – ab) (Adding all the terms)

= ab – bc + bc – ca + ca – ab

= (ab – ab) + (bc – bc) + (ca – ca) (Collecting the like terms together)

= 0 + 0 + 0

= 0

(ii) Given:

a – b + ab, b – c + bc, c – a + ac

We have (a – b + ab) + (b – c + bc) + (c – a + ac) (Adding all the terms)

= a – b + ab + b – c + bc + c – a + ac

= (a – a) + (b – b) + (c – c) + ab + bc + ac (Collecting all the like terms together)

= 0 + 0 + 0 + ab + bc + ac

= ab + bc + ac

(iii) Given:

2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2

By arranging the like terms in the same column, we have

(Adding columnwise)

(iv) Given: l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + nl

By arranging the like terms in the same column, we have

Thus, the sum of the given expressions is 2(l2 + m2 + n2 + lm + mn + nl)

Ex 9.1 Class 8 Maths Question 4.

(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

(6) 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) Arranging the like terms column-wise, we have

[Change the signs of all the terms of lower expressions and then add]

(b) Arranging the like terms column-wise, we have

[Change the signs of all the terms of lower expressions and then add]

(c) Arranging the like terms column-wise, we have

[Change the signs of all the terms of lower expressions and then add]

The terms are p2q – 7pq2 + 8pq – 18q + 5p + 20

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