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**Mathsight Class 6 Solutions Chapter 8 Introduction to Algebra**

Welcome to NCTB Solutions. Here with this post we are going to help 6th class students to solve MathSight Class 6 Mathematics Book, Chapter 8, Introduction to Algebra. Here students can easily get all the exercise questions solution for Chapter 8, Introduction to Algebra Exercise 8.1

**Introduction to Algebra Exercise 8.1 Solution : **

**Question no – (1)**

**Solution :**

**(a)** a – 7

**(b)** 2a/3b

**(c)** (a+b)/2

**(d)** 3a + 10b

**(e)** x/3 (y+z)

**Question no – (2)**

**Solution :**

**(a)** 4x^{2}b^{3}

**(b)** 14 a^{15}

**(c)** 2 a^{7}y^{2}

**(d)** 2p^{5}

**(e)** 6a^{2}b^{3}

**Question no – (3)**

**Solution :**

**(a)** x^{4} = 3 x^{3} = 2 x^{2} = 4 x = 8

**(b)** x^{2} = 3 xy = 4 y^{2} = 1

**(c)** a^{2} = 1 b^{2}c = 3

**(d)** a^{2} = 5 b^{2} = 1

**Question no – (4)**

**Solution : **

**(a)** 5x, 7y, (-2z)

**(b)** 9x^{2}, 3x^{2}y

**(c)** p^{2}q, -pq^{2}, -pq

**(d)** x^{2}, -2x/5

**Question no – (5)**

**Solution :**

**(a)** Coefficient of x is a

**(b)** Coefficient of x is 1

**(c)** Coefficient of x is 0

**(d)** Coefficient of x is 4

**Question no – (6)**

**Solution :**

**(a)** 5 + 4x

**(b)** 3x^{2} – y^{2} + xy

**(c)** 4x^{2} – 7x – 3

**(d)** a^{2} + b^{2} + c^{2} – 3abc

**Question no – (7)**

**Solution :**

**(a)** 4x^{2}y, 3x^{2}y

**(b)** 5a, -2a, -1/3a

**(c)** x^{2}, -2x^{2}

**(d)** 4a^{2}b, a^{2}b, 5a^{2}b

**Question no – (8)**

**Solution :**

**(a)** Trinomial

**(b)** Monomial

**(c)** Monomial

**(d)** Binomial

**(e)** Trinomial

**(f)** Trinomial

**Question no – (9)**

**Solution :**

Algebraic expression in words :

**(a)** 7 in taken any from a

**(b)** 2 time of x increase by 3

**(c)** Sum of y and x is increase by x

**(d)** 10 more the square of t

**Question** **no – (10)**

**Solution :**

**(a)** 2a – 3b + 5

= 2 (-4) – 3 (1) + 5 [∵ a = -4, b = 1]

= -8 – 3 + 5

= – 8 + 2

= – 6

**(b)** x^{2} – 3x^{2}y – 4x + 7

= (-2^{2}) – 3 (-2^{2}) 2 – 4 (-2) + 7 [∵x = -2, y = 2^{2}]

= 4 – 24 + 8 + 7

= – 5

**(c)** x^{2} + y^{2} – z^{2} – 3xyz

= 1^{2} + (-1)^{2} – (2)^{2} – 3.1.(-1).2 [∵ x = 1, y = -1, z = 2]

= 1 + 1 – 4 + 6

= 4

**(d)** 6a + 2b^{2} – ab

= 6 (2) + 2 (3)^{2} – 2.3 [∵ x = 2, y = 3]

= 12 + 18 – 6

= 30 – 6

= 24

**Question no – (11)**

**Solution :**

Present age of tan deep = 5 years

After 10 years of tan deep

= 5 + 10

= 15 years

After 10 years of Akash

= (15 × 2)

= 30 years

∴ Present age of Akash

= 30 – 10

= 20 years

Therefore, the present age of Akash will be 20 years.

**Question no – (12)**

**Solution :**

A = 5 × 5 = 25

**∴** y = 28 – 25 = 3

2 = x + y = x + 3

B = x – y = x – 3

**∴** Area = (x + 3) (x – 3)

= x^{2} – 3x + 3x – 9

= x^{2} – 9

= x^{2} – 3^{2}

Therefore, the area of the rectangle will be x^{2} – 3^{2}

**Previous Chapter Solution : **