Exercise 9.1 of Class 8 Maths Chapter 9 – Algebraic Expressions focuses on identifying terms, coefficients, and variables in algebraic expressions. This exercise helps students understand how expressions are formed using numbers and letters, and lays the groundwork for simplifying and solving algebraic problems in later exercises.
Class 8 Math Ch 9 Algebraic Expression Ex 9.1-NCERT
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) x2 + y2 – 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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