## AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions InText Questions and Answers.

### 8th Class Maths 11th Lesson Algebraic Expressions InText Questions and Answers

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Question 1.

Find the number of terms in following algebraic expressions.

5xy^{2}, 5xy^{3} – 9x, 3xy + 4y – 8, 9x^{2} + 2x + pq + q. [Page No. 248]

Answer:

Question 2.

Take different values for x and find values of 3x + 5. [Page No. 248]

Answer:

If x = 1 then 3x + 5 = 3(1) + 5 = 3 + 5 = 8

If x = 2 then 3x + 5 = 3(2) + 5 = 6 + 5 = 11

If x = 3 then 3x + 5 = 3(3) + 5 = 9 + 5 = 14

Question 3.

Find the like terms in the following: ax^{2}y, 2x, 5y^{2}, -9x^{2}, -6x, 7xy, 18y^{2}. [Pg. No. 249]

Answer:

Like terms are (2x, – 6x) (5y^{2}, 18y^{2}).

Question 4.

Write 3 like terms for 5pq^{2}. [Pg. No. 249]

Answer:

Like terms of 5pq^{2} are – 3pq^{2}, pq^{2}, \(\frac{\mathrm{pq}^{2}}{2}\)etc.,

Question 5.

If A = 2y^{2} + 3x – x^{2}, B = 3x^{2} – y^{2} and C = 5x^{2} – 3xy then find [Pg. No. 250]

(i) A + B (ii) A – B (iii) B + C (iv) B – C (v) A + B + C (vi) A + B – C

Answer:

A = 2y^{2} + 3x – x^{2}, B = 3x^{2} – y^{2}, C = 5x^{2} – 3xy

i) A + B = (2y^{2} + 3x – x^{2}) + (3x^{2} – y^{2})

= (2y^{2} – y^{2}) + 3x + (3x^{2} – x^{2})

∴ A + B = y^{2} + 3x + 2x^{2} = 2x^{2} + 3x + y^{2}

ii) A – B = (2y^{2} + 3x – x^{2}) – (3x^{2} – y^{2})

= 2y^{2} + 3x – x^{2} – 3x^{2} + y^{2}

∴ A – B = 3y^{2} + 3x – 4x^{2}

iii) B + C = (3x^{2} – y^{2}) + (5x^{2} – 3xy)

= 3x^{2} + 5x^{2} – y^{2} – 3xy

∴ B + C = 8x^{2} – y^{2} – 3xy

iv) B – C = (3x^{2} – y^{2}) – (5x^{2} – 3xy)

= 3x^{2} – y^{2} – 5x^{2} + 3xy

∴ B – C = – 2x^{2} – y^{2} + 3xy

v) A + B + C = A + (B + C)

= (2y^{2} + 3x – x^{2}) + (8x^{2} – y^{2} – 3xy)

= (8x^{2} – x^{2}) + (2y^{2} – y^{2}) + 3x – 3xy

∴ A + B + C = 7x^{2} + y^{2} + 3x – 3xy

vi) A + B – C = A + (B – C)

= (2y^{2} + 3x – x^{2}) + (-2x^{2} – y^{2} + 3xy)

= (2y^{2} – y^{2}) + (-x^{2} – 2x^{2}) + 3x + 3xy :

∴ A + B – C = y^{2} – 3x^{2} + 3x + 3xy

Question 6.

Complete the table: [Page No. 253]

Answer:

Question 7.

Check whether you always get a monomial when two monomials are multiplied. [Page No. 253]

Answer:

Yes, the product of two monomials is always a monomial.

Ex: 2xy × 5y = 10xy is a monomial.

Question 8.

Product of two monomials is a monomial? Check. [Pg. No. 253]

Answer:

Yes, the product of two monomials is a monomial.

∵ 2x × y = 2xy

Question 9.

Find the product: (i) 3x(4ax + 8by) (ii) 4a^{2}b(a – 3b) (iii) (p + Sq^{2}) pq (iv) (m^{3} + n^{3}) 5mn^{2} [Pg. No. 255]

Answer:

i) 3x (4ax + 8by) = 3x × 4ax + 3x × 8by

= 12ax^{2} + 24bxy

ii) 4a^{2}b (a – 3b) = 4a^{2}b × a – 4a^{2}b × 3b

= 4a^{3}b – 12a^{2}b^{2}

iii) (p + 3q^{2}) pq = p × pq + 3q^{2} × pq

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

iv) (m^{3} + n^{3}) 5mn^{2} = m^{3} × 5mn^{2} + n^{3} × 5mn^{2}

= 5m^{4}n^{2} + 5mn^{5}

Question 10.

Find the number of maximum terms in the product of a monomial and a binomial? [Pg. No. 255]

Answer:

The no.of terms in the product of a monomial and a binomial are two (2).

Question 11.

Find the product: [Pg. No. 257]

(i) (a – b) (2a + 4b)

(ii) (3x + 2y) (3y – 4x)

(iii) (2m – l)(2l – m)

(iv) (k + 3m)(3m – k)

Answer:

i) (a – b) (2a + 4b) = a(2a + 4b) – b(2a + 4b)

= (a × 2a + a × 4b) – (b × 2a + b × 4b)

= 2a^{2} + 4ab – (2ab + 4b^{2})

= 2a^{2} + 4ab – 2ab – 4b^{2}

= 2a^{2} + 2ab – 4b^{2}

ii) (3x + 2y) (3y – 4x) = 3x(3y – 4x) + 2y(3y – 4x)

= 9xy – 12x^{2} + 6y^{2} – 8xy

= xy – 12x^{2} + 6y^{2}

iii) (2m – l) (2l – m) = 2m(2l – m) – l(2l – m)

= 2m × 2l – 2m × m – l × 2l + l × m

= 4lm – 2m^{2} – 2l^{2} + lm

= 5lm – 2m^{2} – 2l^{2}

iv) (k + 3m) (3m – k) = k(3m – k) + 3m(3m – k)

= k × 3m – k × k + 3m × 3m – 3m × k

= 3km – k^{2} + 9m^{2} – 3km

= 9m^{2} – k^{2}

Question 12.

How many number of terms will be there in the product df two binomials? [Page No. 257]

Answer:

No. of terms in the product of two binomials are 4.

Ex: (a + b) (c + d) = ac + ad + be + bd

Question 13.

Verify the following are identities by taking a, b, c as positive integers. [Pg. No. 260]

(i) (a – b)^{2} = a^{2} – 2ab + b^{2}

(ii) (a + b) (a – b) = a^{2} – b^{2}

(iii) (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

Answer:

i) (a – b)^{2} = a^{2} – 2ab + b^{2}

a = 3, b = 1

⇒ (3 – 1)^{2} = (3)^{2} – 2 × 3 × 1 + 1^{2}

⇒ (2)^{2} = 9 – 6 + 1

⇒ 4 = 4

∴ (i) is an identity,

ii) (a + b) (a – b) = a^{2} – b^{2}

a = 2, b = 1

⇒ (2 + 1) (2 – 1) = (2)^{2} – (1)^{2}

⇒ 3 × 1 = 4 – 1

⇒ 3 = 3

∴ (ii) is an identity.

iii) (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

a = 1, b = 2, c = 0

⇒ (1 + 2 + 0)^{2} = 1^{2} + 2^{2} + 0^{2} + 2 × 1 × 2 + 2 × 2 × 0 + 2 × 0 × 1

⇒ (3)^{2} = 1 + 4 + 0 + 4 + 0 + 0

⇒ 9 = 1 + 4 + 4

⇒ 9 = 9

∴ (iii) is an identity.

Question 14.

Now take x = 2, a = 1 and b = 3, verify the identity (x + a) (x + b) s x + (a + b)x + ab. [Pg. No. 260]

i) What do you observe? Is LHS = RHS?

ii) Take different values for x, a and b for verification of the above identity.

iii) Is it always LHS = RHS for all values of a and b?

Answer:

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

x = 2, a = 1, b = 3 then

⇒ (2 + 1) (2 + 3) = 2^{2} + (1 + 3)^{2} + 1 × 3

⇒ 3 × 5 = 4 + 4x^{2} + 3

⇒ 15 = 4 + 8 + 3 ⇒ 15 = 15

∴ LHS = RHS

ii) x = 0, a = 1, b = 2 then

⇒ (0 + 1) (0 + 2) = 0^{2} + (1 + 2) 0 + 1 × 2

⇒ 1 × 2 = 0 + 0 + 2

⇒ 2 = 2

∴ LHS = RHS for different values of x, a, b.

iii) LHS = RHS for all the values of a, b.

Question 15.

Consider (x + p) (x + q) = x + (p + q)x + pq.

(i) Put q instead of ‘p’ what do you observe?

(ii) Put p instead of ‘q’ what do you observe?

(iii) What identities you observed in your results? [Pg. No. 261]

Answer:

i) (x + p) (x + q) = x^{2} + (p + q)x + pq …… (1)

Substitute q instead of p in (1).

⇒ (x + q) (x + q) = x^{2} + (q + q)x + q × q

⇒ (x + q)^{2} = x^{2} + 2qx + q^{2}

ii) Substitute ‘p’ instead of q in (1).

⇒ (x + p) (x + p) = x^{2} + (p + p)x + p × p

⇒ (x + p) = x^{2} + 2px + p^{2}

iii) ∴ I observe the following identities.

(x + q)^{2} = x^{2} + 2qx + q^{2}

(x + p)^{2} = x^{2} + 2px + p^{2}

Question 16.

Find: (i) (5m + 7n)^{2}

(ii) (6kl + 7mn)^{2}

(iii) (5a^{2} + 6b^{2})^{2}

(iv) 302^{2}

(v) 807^{2}

(vi) 704^{2}

(vii) Verify the identity: (a – b)^{2} = a^{2} – 2ab + b^{2}, where a = 3m and b = 5n. [Pg. No. 261]

Answer:

i) (5m + 7n)^{2} is in the form of (a + b)^{2}.

(a + b)^{2} = a^{2} + 2ab + b^{2} [a = 5m, b = 7n]

(5m + 7n)^{2} = (5m)^{2} + 2 × 5m × 7n + (7n)^{2}

= (5m × 5m) + 70 mn + 7n × 7n

= 25m^{2} + 70mn + 49n^{2}

ii) (6kl + 7mn)^{2}

We know that (a + b)^{2} = a^{2} + 2ab + b^{2}

∴ (6kl + 7mn)^{2} = (6kl)^{2} + 2 × 6kl × 7mn + (7mn)^{2}

= 36 k^{2}l^{2} + 84 klmn + 49 m^{2}n^{2}

iii) (5a^{2} + 6b^{2})^{2}

a = 5a^{2}, b = 6b^{2}

(5a^{2} + 6b^{2})^{2} = (5a^{2})^{2} + 2 × 5a^{2} × 6b^{2} + (6b^{2})^{2}

= (5a^{2} × 5a^{2}) + 60a^{2}b^{2} + (6b^{2} × 6b^{2})

= 25a^{4} + 60a^{2}b^{2} + 36b^{4}

iv) (302)^{2} = (300 + 2)^{2}

a = 300, b = 2

∴ (300 + 2)^{2} = (300)^{2} + 2 × 300 × 2 + (2)^{2}

= (300 × 300) + 1200 + (2 × 2)

= 90,000 + 1200 + 4

= 91,204

v) (807)^{2} = (800 + 7)^{2}

a = 800, b = 7

∴ (800 + 7)^{2} = (800)^{2} + 2 × 800 × 7 + (7)^{2}

= (800 × 800) + 11,200 + (7 × 7)

= 6,40,000 + 11,200 + 49

= 6,51,249

vi) (704)^{2} = (700 + 4)^{2}

a = 700, b = 4

∴ (700 + 4)^{2} = (700)^{2} + 2 × 700 × 4 + 4^{2}

= (700 × 700) + 5600 +(4 × 4)

= 4,90,000 + 5600 + 16

= 4,95,616

vii) (a – b)^{2} = a^{2} – 2ab + b^{2} …… (1)

Substitute a = 3m, b = 5n in (1).

LHS = (3m – 5n)^{2} = (3m)^{2} – 2 × 3m × 5n + (5n)^{2}

= 9m^{2} – 30mn + 25n^{2}

RHS = (3m)^{2} – 2 × 3m × 5n + (5n)^{2}

= 9m^{2} – 30mn + 25n^{2}

∴ LHS = RHS

Question 17.

Find:

(i)(9m – 2n)^{2}

(ii) (6pq – 7rs)^{2}

(iii) (5x^{2} – 6y^{2})^{2}

(iv) 292^{2}

(v) 897^{2}

(vi) 794^{2} [Pg. No. 262]

Answer:

i) (9m – 2n)^{2} is in the form of (a – b)^{2}.

(a – b)^{2} = a^{2} – 2ab + b^{2}

(9m – 2n)^{2} = (9m)^{2} – 2 × 9m × 2n + (2n)^{2}

= (9m × 9m) – 36mn + (2n × 2n)

= 81m^{2} – 36mn + 4n^{2}

ii) (6pq – 7rs)^{2}

a = 6pq, b = 7rs

(6pq – 7rs)^{2} = (6pq)^{2} – 2 × 6pq × 7rs + (7rs)^{2}

= (6pq × 6pq) – 84pqrs + (7rs × 7rs)

= 36p^{2}q^{2} – 84pqrs + 49r^{2}s^{2}

iii) (5x^{2} – 6y^{2})^{2} = (5x^{2})^{2} – 2 × 5x^{2} × 6y^{2} + (6y^{2})^{2}

= (5x^{2} × 5x^{2}) – 60x^{2}y^{2} + (6y^{2} × 6y^{2})

= 25x^{4} – 60x^{2}y^{2} + 36y^{4}

iv) (292)^{2} = (300 – 8)^{2}

a = 300, b = 8

∴ (300 – 8)^{2} = (300)^{2} – 2 × 300 × 8 + (8)^{2} = (300 × 300) – 4800 + (8 × 8)

= 90,000 – 4800 + 64

= 90,064 – 4800

= 85,264

v) (897)^{2} = (900 – 3)^{2}

= (900)^{2} – 2 × 900 × 3 + (3)^{2}

= 8,10,000 – 5400 + 9

= 8,10,009 – 5400

= 8,04,609

vi) (794)^{2} = (800 – 6)^{2}

= (800)^{2} – 2 × 800 × 6 + (6)^{2}

= 6,40,000 – 9600 + 36

= 6,40,036 – 9600

= 6,30,436

Question 18.

Find:

(i) (6m + 7n) (6m – 7n)

(ii) (5a + 10b) (5a – 10b)

(iii) (3x^{2} + 4y^{2}) (3x^{2} – 4y^{2})

(iv) 106 × 94

(v) 592 × 608

(vi) 92^{2} – 8^{2}

(vi) 984^{2} – 16^{2} [Pg. No. 262]

Answer:

i) (6m + 7n) (6m -,7n) is in the form of (a + b) (a – b). (a + b) (a – b) = a^{2} – b^{2},

here a = 6m, b = 7n

(6m + 7n) (6m – 7n) = (6m)^{2} – (7n)^{2}

= 6m × 6m – 7n × 7n

= 36m^{2} – 49n^{2}

ii) (5a + 10b) (5a – 10b) = (5a)^{2} – (10b)^{2} [∵ (a + b) (a – b) = a^{2} – b^{2}]

= 5a × 5a – 10b × 10b

= 25a^{2} – 100b^{2}

iii) (3x^{2} + 4y^{2}) (3x^{2} – 4y^{2})

= (3x^{2})^{2} – (4y^{2})^{2}

= 3x^{2} × 3x^{2} – 4y^{2} × 4y^{2}

= 9x^{4}-16y^{4} [∵ (a + b) (a – b) = a^{2} – b^{2}]

iv) 106 × 94 = (100 + 6) (100 – 6)

= 100^{2} – 6^{2} = 100 × 100 – 6 × 6 [∵ (a + b) (a- b) = a^{2}– b^{2}]

= 10,000 – 36

= 9,964

v) 592 × 608 = (600 – 8) (600 + 8)

= (600)^{2} – (8)^{2}

= 600 × 600 – 8 × 8

= 3,60,000 – 64

= 3,59,936

vi) 92^{2} – 8^{2} is in the form of a^{2} – b^{2} = (a + b) (a – b).

92^{2} – 8^{2} = (92 + 8)(92 – 8)

= 100 × 84

= 8400

vii) 9842 – 162 = (984 + 16) (984 – 16)

= (1000) (968) [∵ (a + b)(a – b) = a^{2} – b^{2}]

= 9,68,000

Try These

Question 1.

Write an algebraic expression using speed and time; simple interest to be paid, using principal and the rate of simple interest. [Pg. No. 251]

Answer:

Distance = speed × time

d = s × t

Question 2.

Can you think of two more such situations, where we can express in algebraic expressions? [Pg. No. 251]

Answer:

Algebraic expressions are used in the following situations:

i) Area of a triangle = \(\frac{1}{2}\) × base × height = \(\frac{1}{2}\) bh

ii) Perimeter of a rectangle = 2(length + breadth) = 2(l + b)

Think, Discuss and Write

Question 1.

Sheela says the sum of 2pq and 4pq is 8p^{2}q^{2} is she right? Give your explanation. [Pg. No. 249]

Answer:

The sum of 2pq and 4pq = 2pq + 4pq = 6pq

According to Sheela’s solution it is 8p^{2}q^{2}.

6pq ≠ 8p^{2}q^{2}

Sheela’s solution is wrong.

Question 2.

Rehman added 4x and 7y and got 1 lxy. Do you agree with Rehman? [Pg. No. 249]

Answer:

The sum of 4x and 7y

= (4x) + (7y)

= 4x + 7y ≠ 11xy

I do not agree with Rehman’s solution.