Chapter 9 Algebraic Expressions and Identities Ex 9.1 Solutions
Question - 1 : - Identify the terms, their coefficients foreach of the following expressions.
(i) 5xyz2 –3zy (ii) 1 + x + x2 (iii) 4x2y2 –4x2y2z2 +z2
(iv) 3– pq + qr – p (v) (x/2) + (y/2) – xy (vi) 0.3a – 0.6ab + 0.5b
Answer - 1 : -
| Sl. No. | Expression | Term | Coefficient |
| i) | 5xyz2 – 3zy | Term: 5xyz2 Term: -3zy | 5 -3 |
| ii) | 1 + x + x2 | Term: 1
Term: x
Term: x2 | 1 1 1 |
| iii) | 4x2y2 – 4x2y2z2 + z2 | Term: 4x2y2
Term: -4 x2y2z2
Term : z2 | 4 -4 1 |
| iv) | 3 – pq + qr – p | Term : 3 -pq qr -p | 3 -1 1 -1 |
| v) | (x/2) + (y/2) – xy | Term : x/2 Y/2 -xy | ½ 1/2 -1 |
| vi) | 0.3a – 0.6ab + 0.5b | Term : 0.3a -0.6ab 0.5b | 0.3 -0.6 0.5 |
Question - 2 : - Classify the following polynomials asmonomials, binomials, trinomials. Which polynomials do not fit in any of thesethree 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
Answer - 2 : -
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.
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
Answer - 3 : -
i) (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 – 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 – 3p2q2 – 3pq + 7pq + 4 + 5
= – p2q2 + 4pq + 9
iv)(l2 + m2) + (m2 + n2) + (n2 + l2) +(2lm + 2mn + 2nl)
= l2 +l2 +m2 +m2 +n2 +n2 +2lm + 2mn + 2nl
= 2l2 +2m2 +2n2 +2lm + 2mn + 2nl
Question - 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
Answer - 4 : -
(a) (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) (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) (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 + 5 pq+ 3pq- 2pq^2 – 5pq^2 + 5 p^2 q –4p^2 q
= 28 + 5p – 18q + 8pq – 7pq2 + p2q