ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.1

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.1

Question 1.
Identify the terms, their numerical as well as literal coefficients in each of the following expressions:
(i) 12x²yz – 4xy²
(ii) 8 + mn + nl – lm
Solution:

Question 2.
Identify monomials, binomials, and trinomials from the following algebraic expressions :
(i) 5p × q × r²
(ii) 3x² + y ÷ 2z
(iii) -3 + 7x²
(iv) \(\frac{5 a^{2}-3 b^{2}+c}{2}\)
(v) 7x⁵ – \(\frac{3 x}{y}\)
(vi) 5p ÷ 3q – 3p² × q²
Solution:
(i) 5p × q × r² = 5pqr²
In this algebraic expression having only one term
therefore it is monomial.

(ii) 3x² + y ÷ 2z = \(\frac{3 x^{2} y}{2 z}\)
In this algebraic expression having only one term
therefore it is monomial.

(iii) -3 + 7x²
In this algebraic expression having two terms
therefore it is binomial.

(iv) \(\frac{5 a^{2}-3 b^{2}+c}{2}=\frac{5 a^{2}}{2}-\frac{3 b^{2}}{2}+\frac{c}{2}\)
In this algebraic expression having three terms
therefore it is trinomial.

(v) 7x⁵ – \(\frac{3 x}{y}\)
In this algebraic expression having two terms
therefore it is binomial.

(vi) 5p ÷ 3q – 3p² × q² = \(\frac{5 p}{3 q}\) = 3p²q²
In this algebraic expression having two terms
therefore it is binomial.

Question 3.
Identify which of the following expressions are polynomials. If so, write their degrees.
(i) \(\frac{2}{5}\)x⁴ – \(\sqrt{3}\)x² + 5x – 1
(ii) 7x³ – \(\frac{3}{x^{2}}\) + \(\sqrt{5}\)
(iii) 4a³b² – 3ab⁴ + 5ab + \(\frac{2}{3}\)
(iv) 2x²y – \(\frac{3}{x y}\) + 5y³ + \(\sqrt{3}\)
Solution:
(i) Polynomial, and degree of this expression is 4.
(ii) Not a polynomial.
(iii) Polynomial, and degree of this expression is 5.
(iv) Not a polynomial.

Question 4.
Add the following expressions:
(i) ab – bv, bv – ca, ca – ab
(ii) 5p²q² + 4pq + 7,3 + 9pq – 2p²q
(iii) l² + m² + n², lm + mn, mn + nl, nl + lm
(iv) 4x³ – 7x² + 9, 3x² – 5x + 4, 7x³ – 11x + 1, 6x² – 13x
Solution:
(i) ab – bc, bc – ca, ca – ab
= ab – bc + bc – ca + ca – ab = 0

(ii) 5p²q² + 4pq + 7,3 + 9pq – 2p²q²
= 5p²q² + 4pq + 7 + 3 + 9pq – 2p²q²
= 5p²q² – 2p²q² + 4pq + 9pq + 7 + 3
= 3p²q² + 13pq + 10

(iii) l² + m² + n², lm + mn, mn + nl, nl + lm
= l² + m² + n² + lm + mn + mn + nl + nl + lm
= l² + m² + n² + 2lm + 2mn + 2nl

(iv) 4x³ – 7x² + 9, 3x² – 5x + 4, 7x³ – 11x + 1, 6x² – 13x
= 4x³ – 7x² + 9 + 3x² – 5x + 4 + 7x³ – 11² + 1 + 6x² – 13x
= 4x² + 7x³ – 7x² + 3x² + 6x² – 5x – 11x – 13x + 9 + 4 + 1
= 11x³ – 2x² – 29x + 14

Question 5.
Subtract:
(i) 8a + 3ab – 2b + 7 from 14a – 5ab + 7b – 5
(ii) 8xy + 4yz + 5zx from 12xy – 3yz – 4zx + 5xyz
(iii) 4p²q – 3pq + 5pq² – 8p + 7q -10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q
Solution:
(i) 8a + 3ab – 2b+ 1 from 14a – 5ab + 7b – 5
= (14a – 5ab + 7b – 5) – (8a + 3ab – 2b + 7)
= 14a – 5ab + 7b – 5 – 8a – 3ab + 2b – 7
= 6a – 8ab + 9ab – 12

(ii) 8xy + 4yz + 5zx from 12xy – 3yz – 4zx + 5xyz
= (12xy – 3yz – 4zx + 5xyz) – (8xy + 4yz + 5zx)
= 12xy – 3yz – 4zx + 5xyz – 8xy – 4yz – 5zx
= 4xy – 7yz – 9zx + 5xyz

(iii) 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 – 7p²q + 3pq – 5pq² + 8p – 7q + 10
= 28 + 5p – 78q + 8pq – 7pq² + p²q

Question 6.
Subtract the sum of 3x² + 5xy + 7y² + 3 and 2x² – 4xy – 3y² + 7 from 9x² – 8xy + 11y²
Solution:
Sum :

Question 7.
What must be subtracted from 3a² – 5ab – 2b² – 3 to get 5a² – 7ab – 3b² + 3a ?
Solution:
We have to subtract 5a² – 7ab – 3b² + 3a
from 3a² – 5ab – 2b² – 3

Question 8.
The perimeter of a triangle is 7p² – 5p + 11 and two of its sides are p² + 2p – 1 and 3p² – 6p + 3. Find the third side of the triangle.
Solution:
We know that,
Perimeter of a triangle = Sum of three sides of triangle
⇒ 7p² – 5p + 11 = (p² + 2p – 1) + (3p² – 6p + 3) + (Third side of triangle)
⇒ 7p² – 5p + 11 = (4p² – 4p + 2) + (Third side of triangle)
⇒ Third side of triangle
= (7p² – 5p + 11) – (4p² – 4p + 2)
= (7p² – 4p²) + (- 5p + 4p) + (11 – 2)
= 3p² – p + 9
Hence third side of triangle is 3p² – p + 9

ML Aggarwal Class 8 Solutions for ICSE Maths