## 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) 12x^{2}yz – 4xy^{2}

(ii) 8 + mn + nl – lm

Solution:

Question 2.

Identify monomials, binomials, and trinomials from the following algebraic expressions :

(i) 5p × q × r^{2}

(ii) 3x^{2} + y ÷ 2z

(iii) -3 + 7x^{2}

(iv) \(\frac{5 a^{2}-3 b^{2}+c}{2}\)

(v) 7x^{5} – \(\frac{3 x}{y}\)

(vi) 5p ÷ 3q – 3p^{2} × q^{2}

Solution:

(i) 5p × q × r^{2} = 5pqr^{2}

In this algebraic expression having only one term

therefore it is monomial.

(ii) 3x^{2} + y ÷ 2z = \(\frac{3 x^{2} y}{2 z}\)

In this algebraic expression having only one term

therefore it is monomial.

(iii) -3 + 7x^{2}

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) 7x^{5} – \(\frac{3 x}{y}\)

In this algebraic expression having two terms

therefore it is binomial.

(vi) 5p ÷ 3q – 3p^{2} × q^{2} = \(\frac{5 p}{3 q}\) = 3p^{2}q^{2}

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) \(\frac{2}{5}\)x^{4} – \(\sqrt{3}\)x^{2} + 5x – 1

(ii) 7x^{3} – \(\frac{3}{x^{2}}\) + \(\sqrt{5}\)

(iii) 4a^{3}b^{2} – 3ab^{4} + 5ab + \(\frac{2}{3}\)

(iv) 2x^{2}y – \(\frac{3}{x y}\) + 5y^{3} + \(\sqrt{3}\)

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) 5p^{2}q^{2} + 4pq + 7,3 + 9pq – 2p^{2}q

(iii) l^{2} + m^{2} + n^{2}, lm + mn, mn + nl, nl + lm

(iv) 4x^{3} – 7x^{2} + 9, 3x^{2} – 5x + 4, 7x^{3} – 11x + 1, 6x^{2} – 13x

Solution:

(i) ab – bc, bc – ca, ca – ab

= ab – bc + bc – ca + ca – ab = 0

(ii) 5p^{2}q^{2} + 4pq + 7,3 + 9pq – 2p^{2}q^{2}

= 5p^{2}q^{2} + 4pq + 7 + 3 + 9pq – 2p^{2}q^{2}

= 5p^{2}q^{2} – 2p^{2}q^{2} + 4pq + 9pq + 7 + 3

= 3p^{2}q^{2} + 13pq + 10

(iii) l^{2} + m^{2} + n^{2}, lm + mn, mn + nl, nl + lm

= l^{2} + m^{2} + n^{2} + lm + mn + mn + nl + nl + lm

= l^{2} + m^{2} + n^{2} + 2lm + 2mn + 2nl

(iv) 4x^{3} – 7x^{2} + 9, 3x^{2} – 5x + 4, 7x^{3} – 11x + 1, 6x^{2} – 13x

= 4x^{3} – 7x^{2} + 9 + 3x^{2} – 5x + 4 + 7x^{3} – 11^{2} + 1 + 6x^{2} – 13x

= 4x^{2} + 7x^{3} – 7x^{2} + 3x^{2} + 6x^{2} – 5x – 11x – 13x + 9 + 4 + 1

= 11x^{3} – 2x^{2} – 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) 4p^{2}q – 3pq + 5pq^{2} – 8p + 7q -10 from 18 – 3p – 11q + 5pq – 2pq^{2} + 5p^{2}q

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) 4p^{2}q – 3pq + 5pq^{2} – 8p + 7q – 10

from 18 – 3p – 11q + 5pq – 2pq^{2} + 5p^{2}q

= (18 – 3p – 11q + 5pq – 2pq^{2} + 5p^{2}q) – (4p^{2}q – 3pq + 5pq^{2} – 8p + 7q – 10)

= 18 – 3p – 11q + 5pq – 2pq^{2} + 5p^{2}q – 7p^{2}q + 3pq – 5pq^{2} + 8p – 7q + 10

= 28 + 5p – 78q + 8pq – 7pq^{2} + p^{2}q

Question 6.

Subtract the sum of 3x^{2} + 5xy + 7y^{2} + 3 and 2x^{2} – 4xy – 3y^{2} + 7 from 9x^{2} – 8xy + 11y^{2}

Solution:

Sum :

Question 7.

What must be subtracted from 3a^{2} – 5ab – 2b^{2} – 3 to get 5a^{2} – 7ab – 3b^{2} + 3a ?

Solution:

We have to subtract 5a^{2} – 7ab – 3b^{2} + 3a

from 3a^{2} – 5ab – 2b^{2} – 3

Question 8.

The perimeter of a triangle is 7p^{2} – 5p + 11 and two of its sides are p^{2} + 2p – 1 and 3p^{2} – 6p + 3. Find the third side of the triangle.

Solution:

We know that,

Perimeter of a triangle = Sum of three sides of triangle

⇒ 7p^{2} – 5p + 11 = (p^{2} + 2p – 1) + (3p^{2} – 6p + 3) + (Third side of triangle)

⇒ 7p^{2} – 5p + 11 = (4p^{2} – 4p + 2) + (Third side of triangle)

⇒ Third side of triangle

= (7p^{2} – 5p + 11) – (4p^{2} – 4p + 2)

= (7p^{2} – 4p^{2}) + (- 5p + 4p) + (11 – 2)

= 3p^{2} – p + 9

Hence third side of triangle is 3p^{2} – p + 9