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

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
x2y – 10x2y + 5xy – 20
Coefficient of x2y = 1
Coefficient of -10x2y = -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, x2y2, 4x3y3 .
(d) a + b + c + d, 3x + 2y + z + 5

Try These (Page 139)

Question 1.
Write two terms which are like
(i) 7xy,
(ii) 4mn2,
(iii) 2l.
Solution:
(i) (7xy + 2); (14xy + 3)
(ii) (4mn2 – 1); (6mn2 + 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) a2(2ab – 5c).
Solution:
(i) 2x(3x + 5xy)
= (2 × 3)x2 + (2 × 5)x2y
= 6x2 + 10x2y.

(ii) a2(2ab – 5c) = 2a3b – 5a2c.

Try These (Page 145)

Question 1.
Find the product (4p2 + 5p + 7) × 3p.
Solution:
(4p2 + 5p + 7) × 3p
= 12p3 + 15p2 + 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 = a2 = a2 + 2ab + b2
If b = -b.
⇒ [a + (-b)]2 = a2 + 2a(-b) + (-b)2
⇒ (a – b)2 = a2 – 2ab + b2
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) = x2 + (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) = x2 + (b + b) x + b × b
= x2 + 2bx + b2.

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)
= x2 + [(-c) + (-c)] x + (-c × -c)
= x2 – 2cx + c2.

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)
= x2 + [a + (-a)]x + a × (-a)
= x2 + (a – a) x – a2
= x2 + 0 – a2
= x2 – a2
= (x + a) (x – a).