Haryana State Board HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities InText Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities InText Questions

Try These (Page 138)

Question 1.

Give five examples of expressions containing one variable and five examples of expression containing two variables.

Solution:

One variable :

(i) x + 3,

(ii) x + 4,

(iii) y + 3,

(iv) 2 + 5,

(v) p + 2.

Two variables :

(i) 2xy + 3,

(ii) 3xy + 4,

(iii) 5xy + 3,

(iv) zx + 9,

(v) 2pq + 3.

Question 2.

Show on the number line x, x – 4, 2x + 1, 3x – 2.

Solution:

Identify the coefficient of each term in the expression

x^{2}y – 10x^{2}y + 5xy – 20

Coefficient of x^{2}y = 1

Coefficient of -10x^{2}y = -10

Coefficient of 5xy = 5.

Question 3.

Classify the following polynomials as monomials, binomials, trinomials

-z + 5, x + y + z, y + z + 100, ab – ac, 17.

Solution:

Monomials : 17

Binomials : -z + 5, ab – ac

Trinomials : x + y + z, y + z + 100.

Question 4.

Construct:

(a) 3 binomials with only x as a variable;

(b) 3 binomials with x and y as variables;

(c) 3 monomials withx and y as variables;

(d) 2 polynomials with 4 or more terms.

Solution:

(a) x + 2, 2x + 5, 3x + 1

(b) 2xy + 3, 3xy + 7, 4xy + 2

(c) xy, x^{2}y^{2}, 4x^{3}y^{3} .

(d) a + b + c + d, 3x + 2y + z + 5

Try These (Page 139)

Question 1.

Write two terms which are like

(i) 7xy,

(ii) 4mn^{2},

(iii) 2l.

Solution:

(i) (7xy + 2); (14xy + 3)

(ii) (4mn^{2} – 1); (6mn^{2} + 2)

(iii) (2l + 5); (4l + 3).

Try These (Page 142)

Question 1.

Can you think of two more such situations, where we may need to multiply algebraic expressions ?

[Hint: (i) Think of Speed and Time, (ii) Think of interest to be paid, the principal and the rate of simple interest etc.]

Solution:

(i) Speed = \(\frac{Distance}{Time}\)

s = \(\frac{d}{t}\)

⇒ d = s × t

(ii) Simple Interest

= \(\frac{Principal × Rate × Time}{100}\)

⇒ S.I. = \(\frac{P × R × T}{100}\)

⇒ P × R × T = S.I. × 100

Try These (Page 143)

Question 1.

Find 4x × 5y × 7z.

First find 4x × 5y and multiply it by 7r; or first find 5y × 7z and multiply it by 4x.

Is the result the same ? What do you observe?

Does the order in which you carry out the multiplication matter ?

Solution:

4x × 5y × 7z = (4x + by) × 8z

= (20xy) × 7z

= 140 xyz.

Try These (Page 144)

Question 1.

Find the product:

(i) 2x(3x + 5xy)

(ii) a^{2}(2ab – 5c).

Solution:

(i) 2x(3x + 5xy)

= (2 × 3)x^{2} + (2 × 5)x^{2}y

= 6x^{2} + 10x^{2}y.

(ii) a^{2}(2ab – 5c) = 2a^{3}b – 5a^{2}c.

Try These (Page 145)

Question 1.

Find the product (4p^{2} + 5p + 7) × 3p.

Solution:

(4p^{2} + 5p + 7) × 3p

= 12p^{3} + 15p^{2} + 21p.

Try These (i) (Page 149)

Question 1.

Put -6 in place of 6 in Identity (I). Do you get Identity (II) ?

Solution:

Identity I

⇒ (a + b)^{2} = a^{2} = a^{2} + 2ab + b^{2}

If b = -b.

⇒ [a + (-b)]^{2} = a^{2} + 2a(-b) + (-b)^{2}

⇒ (a – b)^{2} = a^{2} – 2ab + b^{2}

Hence Identity-II verified.

Try These (ii) (Page 149)

Question 1.

Verify identity (IV), for a = 2, b = 3, x = 5.

Solution:

∵ (x + a) (x + b) = x^{2} + (a + b)x + ab

If, a = 2, b = 3, x = 5

Then (x + a) (x + b)

= (5)^{2} + (2 + 3) × 5 + 2 × 3

= 25 + 25 + 6 = 56.

Question 2.

Consider, the special case of identity (IV) with a = b, what do you get ? Is it related to identity (I) ?

Solution:

If a = b,

Then, (x + a) (x + b) = x^{2} + (b + b) x + b × b

= x^{2} + 2bx + b^{2}.

Question 3.

Consider, the special case of identity (IV) with a = -c and b = -c. What do you get ? Is it related to identity (II) ?

Solution:

If a = -c and b = -c

Then, (x + a) (x + b)

= x^{2} + [(-c) + (-c)] x + (-c × -c)

= x^{2} – 2cx + c^{2}.

Question 4.

Consider the special case of identity (IV) with b = -a. What/ do you get ? Is it related to identity (III) ?

Solution:

If b = -a,

Then, (x + a) (x + b)

= x^{2} + [a + (-a)]x + a × (-a)

= x^{2} + (a – a) x – a^{2}

= x^{2} + 0 – a^{2}

= x^{2} – a^{2}

= (x + a) (x – a).