Rd Sharma 2019 Solutions for Class 8 Math Chapter 6 Algebraic Expressions And Identities are provided here with simple step-by-step explanations. These solutions for Algebraic Expressions And Identities are extremely popular among Class 8 students for Math Algebraic Expressions And Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 Book of Class 8 Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2019 Solutions. All Rd Sharma 2019 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 6.13:

Question 1:

Find each of the following product:
5x2 × 4x3

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices. However, use of these laws are subject to their applicability in the given expressions.

In the present problem, to perform the multiplication, we can proceed as follows:

5x2×4x3=5×4×x2×x3
=20x5                           ( am×an=am+n)

Thus, the answer is 20x5.

Page No 6.13:

Question 2:

Find each of the following product:
−3a2 × 4b4

Answer:

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, am×an=am+n, wherever applicable.

We have:

-3a2×4b4=-3×4×a2×b4=-12a2b4

Thus, the answer is -12a2b4.

Page No 6.13:

Question 3:

Find each of the following product:
(−5xy) × (−3x2yz)

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, am×an=am+n, wherever applicable.

We have:

-5xy×-3x2yz=-5×-3× x×x2×y×y×z=15× x1+2×y1+1×z=15x3y2z

Thus, the answer is 15x3y2z.

Page No 6.13:

Question 4:

Find each of the following product:
14xy×23x2yz2

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the the law of indices, that is, am×an=am+n.

We have:

14xy×23x2yz2=14×23×x×x2×y×y×z2=14×23×x1+2×y1+1×z2=16x3y2z2

Thus, the answer is 16x3y2z2.



Page No 6.14:

Question 5:

Find each of the following product:
-75xy2z×133x2yz2

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+n.

We have:

-75xy2z×133x2yz2=-75×133×x×x2×y2×y×z×z2=-75×133×x1+2×y2+1×z1+2=-9115x3y3x3

Thus, the answer is -9115x3y3x3.
 

Page No 6.14:

Question 6:

Find each of the following product:
-2425x3z×-1516xz2y

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+n.

We have:

-2425x3z×-1516xz2y=-2425×-1516×x3×x×z×z2×y=-2425×-1516×x3+1×z1+2×y
=910x4yz3                                                           

Thus, the answer is 910x4yz3.

Page No 6.14:

Question 7:

Find each of the following product:
-127a2b2×92a3b2c2

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+n.
We have:

-127a2b2×92a3b2c2=-127×92×a2×a3×b2×b2×c2=-127×92×a2+3×b2+2×c2=-16a5b4c2

Thus, the answer is -16a5b4c2.

Page No 6.14:

Question 8:

Find each of the following product:
(-7xy)×14x2yz

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+n.

We have:

-7xy×14x2yz=-7×14×x×x2×y×y×z=-7×14×x1+2×y1+1×z=-74x3y2z

Thus, the answer is -74x3y2z.

Page No 6.14:

Question 9:

Find each of the following product:
(7ab) × (−5ab2c× (6abc2)

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.,  am×an=am+n.

We have:

7ab×-5ab2c×6abc2=7×-5×6×a×a×a×b×b2×b×c×c2=7×-5×6×a1+1+1×b1+2+1×c1+2=-210a3b4c3

Thus, the answer is -210a3b4c3.

Page No 6.14:

Question 10:

Find each of the following product:
(−5a) × (−10a2) × (−2a3)

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+n.
We have:

-5a×-10a2×-2a3=-5×-10×-2×a×a2×a3=-5×-10×-2×a1+2+3=-100a6

Thus, the answer is -100a6.

Page No 6.14:

Question 11:

Find each of the following product:
(−4x2) × (−6xy2) × (−3yz2)

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+n.

We have:

-4x2×-6xy2×-3yz2=-4×-6×-3×x2×x×y2×y×z2=-4×-6×-3×x2+1×y2+1×z2=-72x3y3z2

Thus, the answer is -72x3y3z2.

Page No 6.14:

Question 12:

Find each of the following product:
-27a4×-34a2b×-145b2

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+n.

We have:

-27a4×-34a2b×-145b2=-27×-34×-145×a4×a2×b×b2=-27×34×145×a4+2×b1+2=-27×342×14215×a6×b3=-35a6b3

Thus, the answer is -35a6b3.

Page No 6.14:

Question 13:

Find each of the following product:
79ab2×157ac2b×-35a2c

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+n.

We have:

79ab2×157ac2b×-35a2c=79×157×-35×a×a×a2×b2×b×c2×c=7193×1537×-315×a×a×a2×b2×b×c2×c=7193×15317×-315×a1+1+2×b2+1×c2+1=-a4b3c3

Thus, the answer is -a4b3c3.

Page No 6.14:

Question 14:

Find each of the following product:
43u2vw×-5uvw2×13v2wu

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.,am×an=am+n.

We have:

43u2vw×-5uvw2×13v2wu=43×-5×13×u2×u×u×v×v×v2×w×w2×w=43×-5×13×u2+1+1×v1+1+2×w1+2+1=-209u4v4w4

Thus, the answer is -209u4v4w4.

Page No 6.14:

Question 15:

Find each of the following product:
0.5x×13xy2z4×24x2yz

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+n.

We have:
0.5x×13xy2z4×24x2yz=0.5×13×24×x×x×x2×y2×y×z4×z=0.5×13×24×x1+1+2×y2+1×z4+1=4x4y3z5

Thus, the answer is 4x4y3z5.

Page No 6.14:

Question 16:

Find each of the following product:
43pq2×-14p2r×16p2q2r2

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+n.
We have:

43pq2×-14p2r×16p2q2r2=43×-14×16×p×p2×p2×q2×q2×r×r2=43×-14×16×p1+2+2×q2+2×r1+2=-163p5q4r3

Thus, the answer is -13p5q4r3.

Page No 6.14:

Question 17:

Find each of the following product:
(2.3xy) × (0.1x) × (0.16)

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+n.
We have:
2.3xy×0.1x×0.16=2.3×0.1×0.16×x×x×y=2.3×0.1×0.16×x1+1×y=0.0368x2y

Thus, the answer is 0.0368x2y.

Page No 6.14:

Question 18:

Express each of the following product as a monomials and verify the result in each case for x = 1:
(3x) × (4x) × (−5x)

Answer:

We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.,​ am×an=am+n.
We have:

3x×4x×-5x=3×4×-5×x×x×x=3×4×-5×x1+1+1=-60x3

Substituting x = 1 in LHS, we get:

LHS =3x×4x×-5x=3×1×4×1×-5×1=-60

Putting x = 1 in RHS, we get:
RHS =-60x3=-6013=-60×1=-60


 LHS = RHS for = 1; therefore, the result is correct

Thus, the answer is -60x3.

Page No 6.14:

Question 19:

Express each of the following product as a monomials and verify the result in each case for x = 1:
(4x2) × (−3x) × 45x3

Answer:

We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.,​ am×an=am+n.

We  have:

4x2×-3x×45x3=4×-3×45×x2×x×x3=4×-3×45×x2+1+3=-485x6

 4x2×-3x×45x3=-485x6

Substituting x = 1 in LHS, we get:

LHS=4x2×-3x×45x3=4×12×-3×1×45×13=4×-3×45=-485

Putting x = 1 in RHS, we get:

RHS=-485x6=-485×16=-485

LHS = RHS for = 1; therefore, the result is correct

Thus, the answer is -485x6.

Page No 6.14:

Question 20:

Express each of the following product as a monomials and verify the result in each case for x = 1:
(5x4) × (x2)3 × (2x)2

Answer:

We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., am×an=am+n and amn=amn .

We have:

5x4×x23×2x2 =5x4×x6×22×x2=5×22×x4×x6×x2=5×22×x4+6+2=20x12
5x4×x23×2x2 =20x12

Substituting x = 1 in LHS, we get:

LHS=5x4×x23×2x2 =5×14×123×2×12 =5×1×16×22 =5×1×4=20

Put x =1 in RHS, we get:

RHS =20x12=20×112=20×1=20

 LHS = RHS for = 1; therefore, the result is correct.

Thus, the answer is 20x12.

Page No 6.14:

Question 21:

Express each of the following product as a monomials and verify the result in each case for x = 1:
(x2)3 × (2x) × (−4x) × (5)

Answer:

 We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​am×an=am+n and amn=amn .

We have:

x23×2x×-4x×5=x6×2x×-4x×5=2×-4×5×x6×x×x=2×-4×5×x6+1+1=-40x8

 x23×2x×-4x×5=-40x8

Substituting x = 1 in LHS, we get:​

LHS =x23×2x×-4x×5=123×2×1×-4×1×5=16×2×-4×5=1×2×-4×5=-40

Putting x = 1 in RHS, we get:​

RHS=-40x8=-4018=-40×1=-40

 LHS = RHS for = 1; therefore, the result is correct

Thus, the answer is -40x8.

Page No 6.14:

Question 22:

Write down the product of −8x2y6 and −20xy. Verify the product for x = 2.5, y = 1.

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ am×an=am+n.

We have:

-8x2y6×-20xy=-8×-20×x2×x×y6×y=-8×-20×x2+1×y6+1=-160x3y7
-8x2y6×-20xy=-160x3y7

Substituting x = 2.5 and y = 1 in LHS, we get:

LHS=-8x2y6×-20xy=-82.5216×-202.51=-86.251×-202.51=-50×-50=2500

Substituting x = 2.5 and y = 1 in RHS, we get:​

RHS=-160x3y7=-1602.5317=-16015.625×1=-2500

Because LHS is equal to RHS, the result is correct.

Thus, the answer is -160x3y7.

Page No 6.14:

Question 23:

Evaluate (3.2x6y3) × (2.1x2y2) when x = 1 and y = 0.5

Ans

First multiply the expressions and then substitute the values for the variables.

To multiply algebric experssions use the commutative and the associative laws along with the law of indices, am×an=am+n.
We have,

3.2x6y3×2.1x2y2=3.2×2.1×x6×x2×y3×y2=6.72x8y5
Hence, 3.2x6y3×2.1x2y2=6.72x8y5

Now, substitute 1 for x and 0.5  for y in the result.

6.72x8y5=6.72180.55=6.72×1×0.03125=0.21

Hence, the answer is 0.21.

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ am×an=am+n.

We have:

3.2x6y3×2.1x2y2=3.2×2.1×x6×x2×y3×y2=3.2×2.1×x6+2×y3+2=6.72x8y5

3.2x6y3×2.1x2y2=6.72x8y5

Substituting x = 1 and y = 0.5 in the result, we get:

6.72x8y5=6.72180.55=6.72×1×0.03125=0.21

Thus, the answer is 0.21.

Page No 6.14:

Question 24:

Find the value of (5x6) × (−1.5x2y3) × (−12xy2) when x = 1, y = 0.5.

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ am×an=am+n.

We have:

5x6×-1.5x2y3×-12xy2=5×-1.5×-12×x6×x2×x×y3×y2=5×-1.5×-12×x6+2+1×y3+2=90x9y5

5x6×-1.5x2y3×-12xy2=90x9y5

Substituting x = 1 and y = 0.5 in the result, we get:

90x9y5=90190.55=90×1×0.03125=2.8125

Thus, the answer is 2.8125.

Page No 6.14:

Question 25:

Evaluate (2.3a5b2) × (1.2a2b2) when a = 1 and b = 0.5.

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ am×an=am+n.

We have:

2.3a5b2×1.2a2b2=2.3×1.2×a5×a2×b2×b2=2.3×1.2×a5+2×b2+2=2.76a7b4

2.3a5b2×1.2a2b2=2.76a7b4

Substituting a =1 and b = 0.5 in the result, we get:

2.76a7b4=2.76170.54=2.76×1×0.0625=0.1725

Thus, the answer is 0.1725.

Page No 6.14:

Question 26:

Evaluate (−8x2y6) × (−20xy) for x = 2.5 and y = 1.

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​am×an=am+n.

We have:

-8x2y6×-20xy=-8×-20×x2×x×y6×y=-8×-20×x2+1×y6+1=160x3y7

 -8x2y6×-20xy=160x3y7

Substituting x = 2.5 and y = 1 in the result, we get:

160x3y7=1602.5317=160×15.625=2500

Thus, the answer is 2500.

Page No 6.14:

Question 27:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
(−xy3) × (yx3) × (xy)

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​am×an=am+n.

We have:

-xy3×yx3×xy=-1×x×x3×x×y3×y×y=-1×x1+3+1×y3+1+1=-x5y5

To verify the result, we substitute x = 1 and y = 2 in LHS; we get:

LHS =-xy3×yx3×xy=-1×1×23×2×13×1×2=-1×1×8×2×1×2=-8×2×2=-32

Substituting x = 1 and y = 2 in RHS, we get:​

RHS=-x5y5=-11525=-1×1×32=-32

Because LHS is equal to RHS, the result is correct.

Thus, the answer is -x5y5.

Page No 6.14:

Question 28:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
18x2y4×14x4y2×xy×5

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ am×an=am+n.

We have:

18x2y4×14x4y2×xy×5=18×14×5×x2×x4×x×y4×y2×y=18×14×5×x2+4+1×y4+2+1=532x7y7

To verify the result, we substitute x = 1 and y = 2 in LHS; we get:

LHS=18x2y4×14x4y2×xy×5=18×12×24×14×14×22×1×2×5=18×1×16×14×1×4×1×2×5=2×1×2×5=20

Substituting x = 1 and y = 2 in RHS, we get:​

RHS=532x7y7=5321727=532×1×1284=20

Because LHS is equal to RHS, the result is correct.

Thus, the answer is 532x7y7.

Page No 6.14:

Question 29:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
25a2b×-15b2ac×-12c2

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ am×an=am+n.

We have:

25a2b×-15b2ac×-12c2=25×-15×-12×a2×a×b×b2×c×c2=25×-15×-12×a2+1×b1+2×c1+2=3a3b3c3

 The expression doesn't consist of the variables x and y.

 The result cannot be verified for = 1 and = 2

Thus, the answer is 3a3b3c3.

Page No 6.14:

Question 30:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
-47a2b×-23b2c×-76c2a

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ am×an=am+n.

We have:

-47a2b×-23b2c×-76c2a=-47×-23×-76×a2×a×b×b2×c×c2=-47×-23×-76×a2+1×b1+2×c1+2=-49a3b3c3

 The expression doesn't consist of the variables x and y.

 The result cannot be verified for = 1 and = 2. 

Thus, the answer is -49a3b3c3.

Page No 6.14:

Question 31:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
49abc3×-275a3b2×-8b3c

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ am×an=am+n.

We have:

49abc3×-275a3b2×-8b3c=49×-275×-8×a×a3×b×b2×b3×c3×c=49×-275×-8×a1+3×b1+2+3×c3+1=965a4b6c4

Thus, the answer is 965a4b6c4.​

 The expression doesn't consist of the variables x and y.

 The result cannot be verified for = 1 and = 2

Page No 6.14:

Question 32:

Evaluate each of the following when x = 2, y = −1.
(2xy)×x2y4×x2×y2

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ am×an=am+n.

We have:

2xy×x2y4×x2×y2=2×14×x×x2×x2×y×y×y2=2×14×x1+2+2×y1+1+2=12x5y4

2xy×x2y4×x2×y2=12x5y4

Substituting x = 2 and y = -1 in the result, we get:

12x5y4=1225-14=12×32×1=16

Thus, the answer is 16.

Page No 6.14:

Question 33:

Evaluate each of the following when x = 2, y = −1.
35x2y×-154xy2×79x2y2

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ am×an=am+n.

We have:

35x2y×-154xy2×79x2y2=35×-154×79×x2×x×x2×y×y2×y2=35×-154×79×x2+1+2×y1+2+2=-74x5y5

35x2y×-154xy2×79x2y2=-74x5y5.

Substituting x = 2 and y = -1 in the result, we get:

-74x5y5=-7425-15=-74×32×-1=56

Thus, the answer is 56.



Page No 6.2:

Question 1:

Identify the terms, their coefficients for each of the following expressions:
(i) 7x2yz − 5xy
(ii) x2 + x + 1
(iii) 3x2y2 − 5x2y2z2 + z2
(iv) 9 − ab + bcca
(v) a2+b2-ab
(vi) 0.2x 0.3xy + 0.5y

Answer:

Definitions:

A term in an algebraic expression can be a constant, a variable or a product of constants and variables separated by the signs of addition (+) or subtraction (-) . Examples: 27, x, xyz, 12x2yz etc.
The number factor of the term is called its coefficient.

(i) The expression 7x2yz-5xy consists of two terms, i.e.,  7x2yz and -5xy.
The coefficient of 7x2yz is 7 and the coefficient of -5xy is -5.

(ii) The expression x2+x+1 consists of three terms , i.e.,​ x2, x and 1.
The coefficient of each term is 1.

(iii) The expression 3x2y2-5x2y2z2+z2 consists of three terms , i.e.,​ 3x2y2, -5x2y2z2 and  z2. The coefficient of 3x2y2 is 3. The coefficient of -5x2y2z2 is -5 and the coefficient of z2 is 1.
(iv) The expression 9-ab+bc-ca consists of four terms — 9,-ab, bc and -ca. The coefficient of the term 9 is 9. The coefficient of -ab is -1. The coefficient of bc is 1, and the coefficient of -ca is -1.

(v) The expression a2+b2-ab consists of three terms , i.e.,​ a2, b2 and -ab. The coefficient of a2 is 12. The coefficient of b2 is 12, and the coefficient of -ab is -1.

(vi) The expression 0.2x-0.3xy+0.5y consists of three terms , i.e.,​ 0.2x, -0.3xy and 0.5y. The coefficient of 0.2x is 0.2. The coefficient of -0.3xy is -0.3, and the coefficient of 0.5y is 0.5.

Page No 6.2:

Question 2:

Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any category?
(i) x + y
(ii) 1000
(iii) x + x2 + x3 + 4y4
(iv) 7 + a + 5b
(v) 2b − 3b2
(vi) 2y − 3y2 + 4y3
(vii) 5x − 4y + 3x
(viii) 4a − 15a2
(ix) xy + yz + zt + tx
(x) pqr
(xi) p2q + pq2
(xii) 2p + 2q

Answer:

Definitions:

A polynomial is monomial if it has exactly one term. It is called binomial if it has exactly two non-zero terms. A polynomial ​is a trinomial if it has exactly three non-zero terms.

(i)   The polynomial x+y has exactly two non zero terms , i.e.,​ x and y. Therefore, it is a binomial.
(ii)  The polynomial 1000 has exactly one term, i.e., 1000. Therefore, it is a monomial.
(iii) The polynomial x+x2+x3+x4 has exactly four terms, i.e.,​ x, x2, x3 and x4. Therefore, it doesn't belong to any of the categories.
(iv) The polynomial 7+a+5b has exactly three terms, i.e., 7, and 5b. Therefore, it is a trinomial.
(v)  The polynomial 2b-3b2 has exactly two terms, i.e.,​ 2b and -3b2. Therefore, it is a binomial.
(vi) The polynomial 2y-3y2+4y3 has exactly three terms, i.e.,​ 2y, -3y2 and 4y3. Therefore, it is a trinomial.
(vii) The polynomial 5x-4y+3x has exactly three terms, i.e.,​ 5x, -4y and 3x. Therefore, it is a trinomial.
(viii) The polynomial 4a-15a2 has exactly two terms, i.e., 4a and -15a2. Therefore, it is a binomial.
(ix) The polynomial xy+yz+zt+tx has exactly four terms xy, yz, zt and tx. Therefore, it doesn't belong to any of the categories.
(x) The polynomial pqr has exactly one term, i.e., pqr. Therefore, it is a monomial.
(xi) The polynomial p2q+pq2 has exactly two terms, i.e.,​ p2q and pq2. Therefore, it is a binomial.
(xi) The polynomial 2p+2q has two terms, i.e.,​ 2p and 2q. Therefore, it is a binomial.



Page No 6.21:

Question 1:

Find the following product:
2a3(3a + 5b)

Answer:

To find the product, we will use distributive law as follows:

2a33a+5b=2a3×3a+2a3×5b=2×3a3×a+2×5a3b=2×3a3+1+2×5a3b=6a4+10a3b

Thus, the answer is 6a4+10a3b.

Page No 6.21:

Question 2:

Find the following product:
−11a(3a + 2b)

Answer:

To find the product, we will use distributive law as follows:

-11a3a+2b=-11a×3a+-11a×2b=-11×3×a×a+-11×2×a×b=-33×a1+1+-22×a×b=-33a2-22ab

Thus, the answer is -33a2-22ab.

Page No 6.21:

Question 3:

Find the following product:
−5a(7a − 2b)

Answer:

To find the product, we will use distributive law as follows:

-5a7a-2b=-5a×7a+-5a×-2b=-5×7×a×a+-5×-2×a×b=-35×a1+1+10×a×b=-35a2+10ab

Thus, the answer is -35a2+10ab.

Page No 6.21:

Question 4:

Find the following product:
−11y2(3y + 7)

Answer:

To find the product, we will use distributive law as follows:

-11y23y+7=-11y2×3y+-11y2×7=-11×3y2×y+-11×7×y2=-33y2+1+-77×y2=-33y3-77y2

Thus, the answer is -33y3-77y2.

Page No 6.21:

Question 5:

Find the following product:
6x5(x3+y3)

Answer:

To find the product, we will use distributive law as follows:

6x5x3+y3=6x5×x3+6x5×y3=65×x×x3+65×x×y3=65×x1+3+65×x×y3=6x45+6xy35

Thus, the answer is 6x45+6xy35.

Page No 6.21:

Question 6:

xy(x3y3)

Answer:

To find the product, we will use the distributive law in the following way:

xyx3-y3=xy×x3-xy×y3=x×x3×y-x×y×y3=x1+3y-xy1+3=x4y-xy4

Thus, the answer is x4y-xy4.

Page No 6.21:

Question 7:

Find the following product:
0.1y(0.1x5 + 0.1y)

Answer:

To find the product, we will use distributive law as follows:

0.1y0.1x5+0.1y=0.1y0.1x5+0.1y0.1y=0.1×0.1y×x5+0.1×0.1y×y=0.1×0.1x5×y+0.1×0.1y1+1=0.01x5y+0.01y2

Thus, the answer is 0.01x5y+0.01y2.

Page No 6.21:

Question 8:

Find the following product:
-74ab2c-625a2c2(-50a2b2c2)

Answer:

To find the product, we will use distributive law as follows:

-74ab2c-625a2c2-50a2b2c2=-74ab2c-50a2b2c2-625a2c2-50a2b2c2=-74×-50a×a2×b2×b2×c×c2-625-50a2×a2×b2×c2×c2=-74×-50a1+2b2+2c1+2-625-50a2+2b2c2+2=1752a3b4c3--12a4b2c4=1752a3b4c3+12a4b2c4

Thus, the answer is 1752a3b4c3+12a4b2c4.

Page No 6.21:

Question 9:

Find the following product:
-827xyz32xyz2-94xy2z3

Answer:

To find the product, we will use the distributive law in the following way:

-827xyz32xyz2-94xy2z3=-827xyz32xyz2--827xyz94xy2z3=-827×32x×x×y×y×z×z2--827×94x×x×y×y2×z×z3=-827×32x1+1y1+1z1+2--827×94x1+1y1+2z1+3=-84279×32x1+1y1+1z1+2--82273×94x1+1y1+2z1+3=-49x2y2z3+23x2y3z4

Thus, the answer is -49x2y2z3+23x2y3z4.

Page No 6.21:

Question 10:

Find the following product:
-427xyz92x2yz-34xyz2

Answer:

To find the product, we will use distributive law as follows:

-427xyz92x2yz-34xyz2=-427xyz92x2yz--427xyz34xyz2=-427×92x1+2y1+1z1+1--427×34x1+1y1+1z1+2=-42273×92x1+2y1+1z1+1--41279×34x1+1y1+1z1+2=-23x3y2z2+19x2y2z3

Thus, the answer is -23x3y2z2+19x2y2z3.

Page No 6.21:

Question 11:

Find the following product:
1.5x(10x2y − 100xy2)

Answer:

To find the product, we will use distributive law as follows:

1.5x10x2y-100xy2=1.5x×10x2y-1.5x×100xy2=15x1+2y-150x1+1y2=15x3y-150x2y2

Thus, the answer is 15x3y-150x2y2.

Page No 6.21:

Question 12:

Find the following product:
4.1xy(1.1xy)

Answer:

To find the product, we will use distributive law as follows:

4.1xy1.1x-y=4.1xy×1.1x-4.1xy×y=4.1×1.1×xy×x-4.1xy×y=4.51x1+1y-4.1xy1+1=4.51x2y-4.1xy2

Thus, the answer is 4.51x2y-4.1xy2.

Page No 6.21:

Question 13:

Find the following product:
250.5xyxz+y10

Answer:

To find the product, we will use distributive law as follows:

250.5xyxz+y10=250.5xy×xz+250.5xy×y10=250.5x1+1yz+25.05xy1+1=250.5x2yz+25.05xy2

Thus, the answer is 250.5x2yz+25.05xy2.

Page No 6.21:

Question 14:

Find the following product:
75x2y35xy2+25x

Answer:

To find the product, we will use distributive law as follows:

75x2y35xy2+25x=75x2y×35xy2+75x2y×25x=2125x2+1y1+2+1425x2+1y=2125x3y3+1425x3y

Thus, the answer is 2125x3y3+1425x3y.

Page No 6.21:

Question 15:

Find the following product:
43a(a2 + b2 - 3c2)

Answer:

To find the product, we will use distributive law as follows:

43aa2+b2-3c2=43a×a2+43a×b2-43a×3c2=43a1+2+43ab2-4ac2=43a3+43ab2-4ac2

Thus, the answer is 43a3+43ab2-4ac2.

Page No 6.21:

Question 16:

Find the product 24x2 (1 − 2x) and evaluate its value for x = 3.

Answer:

To find the product, we will use distributive law as follows:

24x21-2x=24x2×1-24x2×2x=24x2-48x1+2=24x2-48x3

Substituting  x = 3 in the result, we get:

24x2-48x3=2432-4833=24×9-48×27=216-1296=-1080

Thus, the product is (24x2-48x3) and its value for x = 3 is (-1080).

Page No 6.21:

Question 17:

Find the product −3y(xy + y2) and find its value for x = 4 and y = 5.

Answer:

To find the product, we will use distributive law as follows:​

-3yxy+y2=-3y×xy+-3y×y2=-3xy1+1-3y1+2=-3xy2-3y3

Substituting x = 4 and y = 5 in the result, we get:

-3xy2-3y3=-3452-353=-3425-3125=-300-375=-675

Thus, the product is (-3xy2-3y3), and its value for ​x = 4 and y = 5 is (-675).

Page No 6.21:

Question 18:

Multiply -32x2y3 by (2x - y) and verify the answer for x = 1 and y = 2.

Answer:

To find the product, we will use distributive law as follows:

-32x2y3×2x-y=-32x2y3×2x--32x2y3×y=-3x2+1y3--32x2y3+1=-3x3y3+32x2y4

Substituting x = 1 and y = 2 in the result, we get:

-3x3y3+32x2y4=-31323+321224=-3×1×8+32×1×16=-24+24=0

Thus, the product is -3x3y3+32x2y4, and its value for ​x = 1 and y = 2 is 0.

Page No 6.21:

Question 19:

Multiply the monomial by the binomial and find the value of each for x = −1, y = 0.25 and z = 0.05:
(i) 15y2(2 − 3x)
(ii) −3x(y2 + z2)
(iii) z2(xy)
(iv) xz(x2 + y2)

Answer:

(i) To find the product, we will use distributive law as follows:

15y22-3x=15y2×2-15y2×3x=30y2-45xy2

Substituting x = -1 and y = 0.25 in the result, we get:

30y2-45xy2=300.252-45-10.252=30×0.0625-45×-1×0.0625=30×0.0625-45×-1×0.0625=1.875--2.8125=1.875+2.8125=4.6875

(ii) To find the product, we will use distributive law as follows:

-3xy2+z2=-3x×y2+-3x×z2=-3xy2-3xz2

Substituting x = -1, y = 0.25​ and z = 0.05​ in the result, we get:

-3xy2-3xz2=-3-10.252-3-10.052=-3-10.0625-3-10.0025=01875+0.0075=0.195

(iii) To find the product, we will use distributive law as follows:

z2x-y=z2×x-z2×y=xz2-yz2

Substituting x = -1, y = 0.25​ and z = 0.05​ in the result, we get:​

xz2-yz2=-10.052-0.250.052=-10.0025-0.250.0025=-0.0025-0.000625=-0.003125

(iv) To find the product, we will use distributive law as follows:

xzx2+y2=xz×x2+xz×y2=x3z+xy2z

Substituting x = -1, y = 0.25​ and z = 0.05​ in the result, we get:​

x3z+xy2z=-130.05+-10.2520.05=-10.05+-10.06250.05=-0.05-0.003125=-0.053125

Page No 6.21:

Question 20:

Simplify:
(i) 2x2(x3x) − 3x(x4 + 2x) − 2(x4 − 3x2)
(ii) x3y(x2 − 2x) + 2xy(x3x4)
(iii) 3a2 + 2(a + 2) − 3a(2a + 1)
(iv) x(x + 4) + 3x(2x2 − 1) + 4x2 + 4
(v) a(b c) − b(ca) − c(ab)
(vi) a(bc) + b(ca) + c(ab)
(vii) 4ab(ab) − 6a2(bb2) − 3b2(2a2a) + 2ab(b a)
(viii) x2(x2 + 1) − x3(x + 1) − x(x3 x)
(ix) 2a2 + 3a(1 − 2a3) + a(a + 1)
(x) a2(2a − 1) + 3a + a3 − 8
(xi) 32x2(x2-1)+14x2(x2+x)-34x(x3-1)
(xii) a2b(ab2) + ab2(4ab − 2a2) − a3b(1 − 2b)
(xiii) a2b(a3a + 1) − ab(a4 − 2a2 + 2a) − b (a3a2 − 1)

Answer:

(i) To simplify, we will use distributive law as follows:

2x2x3-x-3xx4+2x-2x4-3x2=2x5-2x3-3x5-6x2-2x4+6x2=2x5-3x5-2x4-2x3-6x2+6x2=-x5-2x4-2x3

(ii) To simplify, we will use distributive law as follows:​

x3yx2-2x+2xyx3-x4=x5y-2x4y+2x4y-2x5y=x5y-2x5y-2x4y+2x4y=-x5y

(iii) To simplify, we will use distributive law as follows:​

3a2+2a+2-3a2a+1=3a2+2a+4-6a2-3a=3a2-6a2+2a-3a+4=-3a2-a+4

(iv) To simplify, we will use distributive law as follows:

xx+4+3x2x2-1+4x2+4=x2+4x+6x3-3x+4x2+4=x2+4x2+4x-3x+6x3+4=5x2+x+6x3+4

(v) To simplify, we will use distributive law as follows:​

ab-c-bc-a-ca-b=ab-ac-bc+ba-ca+cb=ab+ba-ac-ca-bc+cb=2ab-2ac

(vi) To simplify, we will use distributive law as follows:​

ab-c+bc-a+ca-b=ab-ac+bc-ba+ca-cb   =ab-ba-ac+ca+bc-cb=0

(vii) To simplify, we will use distributive law as follows:​

4aba-b-6a2b-b2-3b22a2-a+2abb-a=4a2b-4ab2-6a2b+6a2b2-6b2a2+3b2a+2ab2-2a2b=4a2b-6a2b-2a2b-4ab2+3b2a+2ab2+6a2b2-6b2a2=-4a2b+ab2

(viii) To simplify, we will use distributive law as follows:​

x2x2+1-x3x+1-xx3-x=x4+x2-x4-x3-x4+x2=x4-x4-x4-x3+x2+x2=-x4-x3+2x2

(ix) To simplify, we will use distributive law as follows:​

2a2+3a1-2a3+aa+1=2a2+3a-6a4+a2+a=2a2+a2+3a+a-6a4=3a2+4a-6a4

(x) To simplify, we will use distributive law as follows:​

a22a-1+3a+a3-8=2a3-a2+3a+a3-8=2a3+a3-a2+3a-8=3a3-a2+3a-8

(xi) To simplify, we will use distributive law as follows:​

32x2x2-1+14x2x2+x-34xx3-1=32x4-32x2+14x4+14x3-34x4+34x=32x4+14x4-34x4+14x3-32x2+34x=6+1-34x4+14x3-32x2+34x=x4+14x3-32x2+34x

(xii) To simplify, we will use distributive law as follows:

a2ba-b2+ab24ab-2a2-a3b1-2b=a3b-a2b3+4a2b3-2a3b2-a3b+2a3b2=a3b-a3b-a2b3+4a2b3-2a3b2+2a3b2=3a2b3

(xiii) To simplify, we will use distributive law as follows:​

a2ba3-a+1-aba4-2a2+2a-ba3-a2-1=a5b-a3b+a2b-a5b+2a3b-2a2b-a3b+a2b+b=a5b-a5b-a3b+2a3b-a3b+a2b-2a2b+a2b+b=b



Page No 6.30:

Question 1:

Multiply:
(5x + 3) by (7x + 2)

Answer:

To multiply, we will use distributive law as follows:

5x+37x+2=5x7x+2+37x+2=5x×7x+5x×2+3×7x+3×2=35x2+10x+21x+6=35x2+10x+21x+6=35x2+31x+6

Thus, the answer is 35x2+31x+6.

Page No 6.30:

Question 2:

Multiply:
(2x + 8) by (x − 3)

Answer:

To multiply the expressions, we will use the distributive law in the following way:

2x+8x-3=2xx-3+8x-3=2x×x-2x×3+8x-8×3=2x2-6x+8x-24=2x2-6x+8x-24=2x2+2x-24

Thus, the answer is 2x2+2x-24.

Page No 6.30:

Question 3:

Multiply:
(7x + y) by (x + 5y)

Answer:

To multiply, we will use distributive law as follows:

7x+yx+5y=7xx+5y+yx+5y=7x2+35xy+xy+5y2=7x2+36xy+5y2

Thus, the answer is 7x2+36xy+5y2.

Page No 6.30:

Question 4:

Multiply:
(a − 1) by (0.1a2 + 3)

Answer:

To multiply, we will use distributive law as follows:

a-10.1a2+3=0.1a2a-1+3a-1=0.1a3-0.1a2+3a-3

Thus, the answer is 0.1a3-0.1a2+3a-3.

Page No 6.30:

Question 5:

Multiply:
(3x2 + y2) by (2x2 + 3y2)

Answer:

To multiply, we will use distributive law as follows:

3x2+y22x2+3y2=3x22x2+3y2+y22x2+3y2=6x4+9x2y2+2x2y2+3y4=6x4+11x2y2+3y4

Thus, the answer is 6x4+11x2y2+3y4.

Page No 6.30:

Question 6:

Multiply:
35x+12y by 56x+4y

Answer:

To multiply, we will use distributive law as follows:

35x+12y56x+4y=35x56x+4y+12y56x+4y=12x2+125xy+512xy+2y2=12x2+144+2560xy+2y2=12x2+16960xy+2y2

Thus, the answer is 12x2+16960xy+2y2.



Page No 6.31:

Question 7:

Multiply:
(x6y6) by (x2 + y2)

Answer:

To multiply, we will use distributive law as follows:

x6-y6x2+y2=x6x2+y2-y6x2+y2=x8+x6y2-y6x2+y8=x8+x6y2-y6x2-y8

Thus, the answer is x8+x6y2-y6x2-y8.

Page No 6.31:

Question 8:

Multiply:
(x2 + y2) by (3a + 2b)

Answer:

To multiply, we will use distributive law as follows:

x2+y23a+2b=x23a+2b+y23a+2b=3ax2+2bx2+3ay2+2by2

Thus, the answer is 3ax2+2bx2+3ay2+2by2.

Page No 6.31:

Question 9:

Multiply:
[−3d + (−7f)] by (5d + f)

Answer:

To multiply, we will use distributive law as follows:

-3d+-7f5d+f=-3d5d+f+-7f5d+f=-15d2-3df+-35df-7f2=-15d2-3df-35df-7f2=-15d2-38df-7f2

Thus, the answer is -15d2-38df-7f2.

Page No 6.31:

Question 10:

Multiply:
(0.8a − 0.5b) by (1.5a − 3b)

Answer:

To multiply, we will use distributive law as follows:

0.8a-0.5b1.5a-3b=0.8a1.5a-3b-0.5b1.5a-3b=1.2a2-2.4ab-0.75ab+1.5b2=1.2a2-3.15ab+1.5b2

Thus, the answer is 1.2a2-3.15ab+1.5b2.

Page No 6.31:

Question 11:

Multiply:
(2x2y2 − 5xy2) by (x2y2)

Answer:

To multiply, we will use distributive law as follows:

2x2y2-5xy2x2-y2=2x2y2x2-y2-5xy2x2-y2=2x4y2-2x2y4-5x3y2+5xy4

Thus, the answer is 2x4y2-2x2y4-5x3y2+5xy4.

Page No 6.31:

Question 12:

Multiply:
x7+x22by25+9x4

Answer:

To multiply the expressions, we will use the distributive law in the following way:

x7+x2225+9x4=x725+9x4+x2225+9x4=2x35+9x228+x25+9x38=2x35+45+28140x2+ 9x38=2x35+73x2140 +9x28
Thus, the answer is  2x35+73x2140+9x38

Page No 6.31:

Question 13:

Multiply:
-a7+a29byb2-b23

Answer:

To multiply, we will use distributive law as follows:

-a7+a29b2-b23=-a7b2-b23+a29b2-b23=-ab14+ab221+a2b18-a2b227=-ab14+ab221+a2b18-a2b227

Thus, the answer is -ab14+ab221+a2b18-a2b227.

Page No 6.31:

Question 14:

Multiply:
(3x2y − 5xy2) by 15x2 + 13y2

Answer:

To multiply, we will use distributive law as follows:

3x2y-5xy215x2+13y2=15x23x2y-5xy2+13y23x2y-5xy2=35x4y-x3y2+x2y3-53xy4

Thus, the answer is 35x4y-x3y2+x2y3-53xy4.

Page No 6.31:

Question 15:

Multiply:
(2x2 − 1) by (4x3 + 5x2)

Answer:

To multiply, we will use distributive law as follows:

2x2-14x3+5x2=2x24x3+5x2-14x3+5x2=8x5+10x4-4x3-5x2

Thus, the answer is 8x5+10x4-4x3-5x2.

Page No 6.31:

Question 16:

(2xy + 3y2) (3y2 − 2)

Answer:

To multiply, we will use distributive law as follows:

2xy+3y23y2-2=2xy3y2-2+3y23y2-2=6xy3-4xy+9y4-6y2=9y4+6xy3-6y2-4xy

Thus, the answer is 9y4+6xy3-6y2-4xy.

Page No 6.31:

Question 17:

Find the following product and verify the result for x = − 1, y = − 2:
(3x − 5y) (x + y)

Answer:

To multiply, we will use distributive law as follows:

3x-5yx+y=3xx+y-5yx+y=3x2+3xy-5xy-5y2=3x2-2xy-5y2

3x-5yx+y=3x2-2xy-5y2.
Now, we put x = -1 and y = -2 on both sides to verify the result.

LHS=3x-5yx+y=3-1-5-2-1+-2=-3+10-3=7-3=-21

RHS=3x2-2xy-5y2=3-12-2-1-2-5-22=3×1-4-5×4=3-4-20=-21

Because LHS is equal to RHS, the result is verified.

Thus, the answer is 3x2-2xy-5y2.

Page No 6.31:

Question 18:

Find the following product and verify the result for x = − 1, y = − 2:
(x2y  1) (3  2x2y)

Answer:

To multiply, we will  use distributive law as follows:

x2y-13-2x2y=x2y3-2x2y-1×3-2x2y=3x2y-2x4y2-3+2x2y=5x2y-2x4y2-3

x2y-13-2x2y=5x2y-2x4y2-3

Now, we put x = -1 and y = -2 on both sides to verify the result.

LHS = x2y-13-2x2y=-12-2-13-2-12-2=1×-2-13-2×1×-2=-2-13+4=-3×7=-21

RHS=5x2y-2x4y2-3=5-12-2-2-14-22-3=5×1×-2-2×1×4-3=-10-8-3=-21

Because LHS is equal to RHS, the result is verified.

Thus, the answer is 5x2y-2x4y2-3.

Page No 6.31:

Question 19:

Find the following product and verify the result for x = − 1, y = − 2:
13x-y2513x+y25

Answer:

To multiply, we will use distributive law as follows:

13x-y2513x+y25=13x13x+y25-y2513x+y25=19x2+xy215-xy215+y425=19x2+xy215-xy215-y425=19x2-y425

 13x-y2513x+y25=19x2-y425

Now, we will put x = -1 and y = -2 on both the sides to verify the result.

LHS = 13x-y2513x+y25=13-1--22513-1+-225=-13-45-13+45=-1715715=-119225

RHS=19x2-y425=19-12--2425=19×1-1625=19-1625=-119225

Because LHS is equal to RHS, the result is verified.

Thus, the answer is 19x2-y425.

Page No 6.31:

Question 20:

Simplify:
x2(x + 2y) (x − 3y)

Answer:

To simplify, we will proceed as follows:

x2x+2yx-3y=x2x+2yx-3y=x3+2x2yx-3y=x3x-3y+2x2yx-3y=x4-3x3y+2x3y-6x2y2=x4-x3y-6x2y2

Thus, the answer is x4-x3y-6x2y2.

Page No 6.31:

Question 21:

Simplify:
(x2 − 2y2) (x + 4y) x2y2

Answer:

To simplify, we will proceed as follows:

x2-2y2x+4yx2y2=x2x+4y-2y2x+4yx2y2=x3+4x2y-2xy2-8y3x2y2=x5y2+4x4y3-2x3y4-8x2y5

Thus, the answer is x5y2+4x4y3-2x3y4-8x2y5.

Page No 6.31:

Question 22:

Simplify:
a2b2(a + 2b)(3a + b)

Answer:

To simplify, we will proceed as follows:

a2b2a+2b3a+b=a2b2a+2b3a+b=a3b2+2a2b33a+b=3aa3b2+2a2b3+ba3b2+2a2b3=3a4b2+6a3b3+a3b3+2a2b4=3a4b2+7a3b3+2a2b4

Thus, the answer is 3a4b2+7a3b3+2a2b4.

Page No 6.31:

Question 23:

Simplify:
x2(x − y) y2(x + 2y)

Answer:

To simplify, we will proceed as follows:

x2x-yy2x+2y=x2x-yy2x+2y=x3-x2yxy2+2y3=x3xy2+2y3-x2yxy2+2y3=x4y2+2x3y3-x3y3+2x2y4=x4y2+2x3y3-x3y3-2x2y4=x4y2+x3y3-2x2y4

Thus, the answer is x4y2+x3y3-2x2y4.

Page No 6.31:

Question 24:

Simplify:
(x3 − 2x2 + 5x − 7)(2x − 3)

Answer:

To simplify, we will proceed as follows:

x3-2x2+5x-72x-3=2xx3-2x2+5x-7-3x3-2x2+5x-7=2x4-4x3+10x2-14x-3x3+6x2-15x+21
=2x4-4x3-3x3+10x2+6x2-14x-15x+21     (Rearranging)
=2x4-7x3+16x2-29x+21                              (Combining like terms)

Thus, the answer is 2x4-7x3+16x2-29x+21.

Page No 6.31:

Question 25:

Simplify:
(5x + 3)(x − 1)(3x − 2)

Answer:

To simplify, we will proceed as follows:

5x+3x-13x-2=5x+3x-13x-2
=5xx-1+3x-13x-2              (Distributive law)
=5x2-5x+3x-33x-2=5x2-2x-33x-2=3x5x2-2x-3-25x2-2x-3=15x3-6x2-9x-10x2-4x-6=15x3-6x2-9x-10x2+4x+6
=15x3-6x2-10x2-9x+4x+6              (Rearranging)
=15x3-16x2-5x+6                              (Combining like terms)

Thus, the answer is 15x3-16x2-5x+6.

Page No 6.31:

Question 26:

Simplify:
(5 − x)(6 − 5x)( 2 − x)

Answer:

To simplify, we will proceed as follows:

5-x6-5x2-x=5-x6-5x2-x
=56-5x-x6-5x2-x                (Distributive law)
=30-25x-6x+5x22-x=30-31x+5x22-x=230-31x+5x2-x30-31x+5x2=60-62x+10x2-30x+31x2-5x3
=60-62x-30x+10x2+31x2-5x3              (Rearranging)
=60-92x+41x2-5x3                                (Combining like terms)

Thus, the answer is 60-92x+41x2-5x3.

Page No 6.31:

Question 27:

Simplify:
(2x2 + 3x − 5)(3x2 − 5x + 4)

Answer:

To simplify, we will proceed as follows:

2x2+3x-53x2-5x+4
=2x23x2-5x+4+3x3x2-5x+4-53x2-5x+4           (Distributive law)
=6x4-10x3+8x2+9x3-15x2+12x-15x2+25x-20
=6x4-10x3+9x3+8x2-15x2-15x2+12x+25x-20              (Rearranging)
=6x4-x3-22x2+36x-20                                                     (Combining like terms)

Thus, the answer is 6x4-x3-22x2+36x-20.

Page No 6.31:

Question 28:

Simplify:
(3x − 2)(2x − 3) + (5x − 3)(x + 1)

Answer:

To simplify, we will proceed as follows:

3x-22x-3+5x-3x+1=3x-22x-3+5x-3x+1
=3x2x-3-22x-3+5xx+1-3x+1           (Distributive law)
=6x2-9x-4x+6+5x2+5x-3x-3
=6x2+5x2-9x-4x+5x-3x-3+6                                  (Rearranging)
=11x2-11x+3                                                                 (Combining like terms)

Thus, the answer is 11x2-11x+3.

Page No 6.31:

Question 29:

Simplify:
(5x − 3)(x + 2) − (2x + 5)(4x − 3)

Answer:

To simplify, we will proceed as follows:

5x-3x+2-2x+54x-3=5x-3x+2-2x+54x-3
=5xx+2-3x+2-2x4x-3+54x-3            (Distributive law)
=5x2+10x-3x-6-8x2+6x-20x+15
=5x2-8x2+10x-3x+6x-20x-6+15                              (Rearranging)
=5x2-8x2+10x-3x+6x-20x-6+15=-3x2-7x+9                              (Combining like terms)

Hence, the answer is -3x2-7x+9.

Page No 6.31:

Question 30:

Simplify:
(3x + 2y)(4x + 3y) − (2xy)(7x − 3y)

Answer:

To simplify, we will proceed as follows:

3x+2y4x+3y-2x-y7x-3y=3x+2y4x+3y-2x-y7x-3y
=3x4x+3y+2y4x+3y-2x7x-3y-y7x-3y            (Distributive law)
=12x2+9xy+8xy+6y2-14x2-6xy-7xy+3y2=12x2+9xy+8xy+6y2-14x2+6xy+7xy-3y2
=12x2-14x2+9xy+8xy+6xy+7xy+6y2-3y2                              (Rearranging)
=-2x2+30xy+3y2                                                                       (Combining like terms)

Thus, the answer is -2x2+30xy+3y2.

Page No 6.31:

Question 31:

Simplify:
(x2 − 3x + 2)(5x − 2) − (3x2 + 4x − 5)(2x − 1)

Answer:

To simplify, we will  proceed as follows:

x2-3x+25x-2-3x2+4x-52x-1=x2-3x+25x-2-3x2+4x-52x-1
=5xx2-3x+2-2x2-3x+2-2x3x2+4x-5-1×3x2+4x-5            (Distributive law)
=5x3-15x2+10x-2x2-6x+4-6x3+8x2-10x-3x2-4x+5=5x3-15x2+10x-2x2+6x-4-6x3+8x2-10x-3x2-4x+5=5x3-15x2+10x-2x2+6x-4-6x3-8x2+10x+3x2+4x-5
=5x3-6x3-15x2-2x2-8x2+3x2+10x+6x+10x+4x-5-4                                (Rearranging)
=-x3-22x2+30x-9                                                                                            (Combining like terms)

Thus, the answer is -x3-22x2+30x-9.
 

Page No 6.31:

Question 32:

Simplify:
(x3 − 2x2 + 3x − 4) (x −1) − (2x − 3)(x2x + 1)

Answer:

To simplify,we will proceed as follows:

x3-2x2+3x-4x-1-2x-3x2-x+1=x3-2x2+3x-4x-1-2x-3x2-x+1
=xx3-2x2+3x-4-1x3-2x2+3x-4-2xx2-x+1-3x2-x+1            (Distributive law)
=xx3-2x2+3x-4-1x3-2x2+3x-4-2xx2-x+1-3x2-x+1=x4-2x3+3x2-4x-x3+2x2-3x+4-2x3-2x2+2x-3x2+3x-3=x4-2x3+3x2-4x-x3+2x2-3x+4-2x3+2x2-2x+3x2-3x+3
=x4-2x3-2x3-x3+3x2+2x2+2x2+3x2-4x-3x-2x-3x+4+3                             (Rearranging)
=x4-5x3+10x2-12x+7                                                                                            (Combining like terms)

Thus, the answer is x4-5x3+10x2-12x+7.



Page No 6.43:

Question 1:

Write the following squares of binomials as trinomials:
(i) (x + 2)2
(ii) (8a + 3b)2
(iii) (2m + 1)2
(iv) 9a+162
(v) x+x222
(vi) x4-y3
(vii) 3x-13x2
(viii) xy-yx2
(ix) 3a2-5b42
(x) (a2bbc2)2
(xi) 2a3b+2b3a2
(xii) (x2ay)2

Answer:

 We will use the identities a+b2=a2+2ab+b2 and a-b2=a2-2ab+b2 to convert the squares of binomials as trinomials.

(i) x+22=x2+2×x×2+b2=x2+4x+b2

(ii) 8a+3b2=8a2+28a3b+6b2=64a2+48ab+36b2

(iii) 2m+12=2m2+22m1+12=4m2+4m+1

(iv) 9a+162=9a2+29a16+162=81a2+3a+136

(v) x+x222=x2+2xx22+x222=x2+x3+x44

(vi) x4-y32=x42-2x4y3+y32=x216-16xy+y29


(vii) 3x-13x2=3x2-23x13x+13x2=9x2-2+19x2

(viii) xy-yx2=xy2-2xyyx+yx2=x2y2-2+y2x2


(ix) 3a2-5b42=3a22-23a25b4+5b42=9a24-15ab4+25b216

(x) a2b-bc22=a2b2-2a2bbc2+bc22=a4b2-2a2b2c2+b2c4




(xi) 2a3b+2b3a2=2a3b2+22a3b2b3a+2b3a2=4a29b2+89+4b29a2



(xii) x2-ay2=x22-2x2ay+ay2=x4-2x2ay+a2y2

Page No 6.43:

Question 2:

Find the product of the following binomials:
(i) (2x + y)(2x + y)
(ii) (a + 2b)(a − 2b)
(iii) (a2 + bc)(a2 bc)
(iv) 4x5-3y44x5+3y4
(v) 2x+3y2x-3y
(vi) (2a3 + b3)(2a3b3)
(vii) x4+2x2x4-2x2
(viii) x3+1x3x3-1x3

Answer:

(i) We will use the identity a+b2=a2+2ab+b2  in the given expression to find the product.
2x+y2x+y=2x+y2=2x2+22xy+y2=4x2+4xy+y2

(ii) We will use the identity a+ba-b=a2-b2 in the given expression to find the product.
a+2ba-2b=a2-2b2=a2-4b2

(iii) We will use the identity a+ba-b=a2-b2 in the given expression to find the product.
a2+bca2-bc=a22-bc2=a4-b2c2

(iv)We will use the identity a+ba-b=a2-b2 in the given expression to find the product.
4x5-3y44x5+3y4=4x52-3y42=16x225-9y216

(v) We will use the identity a+ba-b=a2-b2 in the given expression to find the product.
2x+3y2x-3y=2x2-3y2=4x2-9y2

(vi) We will use the identity a+ba-b=a2-b2 in the given expression to find the product.
2a3+b32a3-b3=2a32-b32=4a6-b6

(vii) We will use the identity a+ba-b=a2-b2 in the given expression to find the product.
x4+2x2x4-2x2=x42-2x22=x8-4x4

(viii) We will use the identity a+ba-b=a2-b2 in the given expression to find the product.

x3+1x3x3-1x3=x32-1x32=x6-1x6

Page No 6.43:

Question 3:

Using the formula for squaring a binomial, evaluate the following:
(i) (102)2
(ii) (99)2
(iii) (1001)2
(iv) (999)2
(v) (703)2

Answer:

(i) Here, we will use the identity a+b2=a2+2ab+b2
1022=100+22=1002+2×100×2+22=10000+400+4=10404

(ii) Here, we will use the identity a-b2=a2-2ab+b2
992=100-12=1002-2×100×1+12=10000-200+1=9801

(iii) Here, we will use the identity a+b2=a2+2ab+b2
10012=1000+12=10002+2×1000×1+12=1000000+2000+1=1002001

(iv) Here, we will use the identity a-b2=a2-2ab+b2
9992=1000-12=10002-2×1000×1+12=1000000-2000+1=998001

(v) Here, we will use the identity a+b2=a2+2ab+b2
7032=700+32=7002+2×700×3+32=490000+4200+9=494209

Page No 6.43:

Question 4:

Simplify the following using the formula: (ab)(a + b) = a2b2:
(i) (82)2 − (18)2
(ii) (467)2 − (33)2
(iii) (79)2 − (69)2
(iv) 197 × 203
(v) 113 × 87
(vi) 95 × 105
(vii) 1.8 × 2.2
(viii) 9.8 × 10.2

Answer:

Here, we will use the identity (a-b)(a+b)=a2 -b2

(i) Let us consider the following expression:

822-182=82+1882-18=100×64=6400

(ii) Let us consider the following expression:

4672-332=467+33467-33=500×434=217000

(iii) Let us consider the following expression:

792-692=79+6979-69=148×10=1480

(iv) Let us consider the following product:

197×203

 197+2032=4002=200; therefore, we will write the above product as:
197×203=200-3200+3=2002-32=40000-9=39991

Thus, the answer is 39991.

(v) Let us consider the following product:

113×87

113+872=2002=100; therefore, we will write the above product as:
113×87=100+13100-13=1002-132=10000-169=9831

Thus, the answer is 9831.

(vi) Let us consider the following product:

95×105

95+1052=2002=100; therefore, we will write the above product as:
95×105=100+5100-5=1002-52=10000-25=9975

Thus, the answer is 9975.

(vii) Let us consider the following product:

1.8×2.2

1.8+2.22=42=2; therefore, we will write the above product as:
1.8×2.2=2-0.22+0.2=22-0.22=4-0.04=3.96

Thus, the answer is 3.96.

(viii) Let us consider the following product:

9.8×10.2

9.8+10.22=202=10; therefore, we will write the above product as:
9.8×10.2=10-0.210+0.2=102-0.22=100-0.04=99.96

Thus, the answer is 99.96.

Page No 6.43:

Question 5:

Simplify the following using the identities:
(i) 582-42216
(ii) 178 × 178 − 22 × 22
(iii) 198×198-102×10296
(iv) 1.73 × 1.73 − 0.27 × 0.27
(v) 8.63×8.63-1.37×1.370.726

Answer:

(i) Let us consider the following expression:

582-42216
Using the identity a+ba-b=a2-b2, we get:

582-42216=58+4258-4216
582-42216=100×1616582-42216=100

Thus, the answer is 100.

(ii) Let us consider the following expression:

178×178-22×22
Using the identity a+ba-b=a2-b2, we get:

178×178-22×22=1782-222=178+22178-22=200×156=31200
Thus, the answer is 31200.

(iii) Let us consider the following expression:

198×198-102×10296=1982-102296
Using the identity a+ba-b=a2-b2, we get:

198×198-102×10296=1982-102296=198+102198-10296
198×198-102×10296=198+102198-10296198×198-102×10296=300×9696198×198-102×10296=300

Thus, the answer is 300.

(iv) Let us consider the following expression:

1.73×1.73-0.27×0.27
Using the identity a+ba-b=a2-b2, we get:

1.73×1.73-0.27×0.27=1.732-0.272=1.73+0.271.73-0.27=2×1.46=2.92
Thus, the answer is 2.92.

(v) Let us consider the following expression:

8.63×8.63-1.37×1.370.726=8.632-1.3720.726
Using the identity a+ba-b=a2-b2, we get:

8.63×8.63-1.37×1.370.726=8.632-1.3720.726=8.63+1.378.63-1.370.726
8.63×8.63-1.37×1.370.726=8.63+1.378.63-1.370.7268.63×8.63-1.37×1.370.726=8.63+1.378.63-1.370.7268.63×8.63-1.37×1.370.726=10×7.260.7268.63×8.63-1.37×1.370.726=10×7.26100.7268.63×8.63-1.37×1.370.726=100

Thus, the answer is 100.

Page No 6.43:

Question 6:

Find the value of x, if:
(i) 4x = (52)2 − (48)2
(ii) 14x = (47)2 − (33)2
(iii) 5x = (50)2 − (40)2

Answer:

(i) Let us consider the following equation:

4x=522-482
Using the identity a+ba-b=a2-b2, we get:
4x=522-4824x=52+4852-484x=100×4=400
4x=400
x=100        (Dividing both sides by 4)

(ii) Let us consider the following equation:

14x=472-332
Using the identity a+ba-b=a2-b2, we get:
14x=472-33214x=47+3347-3314x=80×14=1120
14x=1120
x=80        (Dividing both sides by 14)

(iii) Let us consider the following equation:

5x=502-402
Using the identity a+ba-b=a2-b2, we get:
5x=502-4025x=50+4050-405x=90×10=900
5x=900
x=180        (Dividing both sides by 5)

Page No 6.43:

Question 7:

If x+1x=20, find the value of x2+1x2.

Answer:

Let us consider the following equation:

x+1x=20
Squaring both sides, we get:

x+1x2=202=400x+1x2=400x2+2×x×1x+1x2=400                   [(a+b)2=a2 +b2 +2ab]x2+2+1x2=400
x2+1x2=398                               (Subtracting 2 from both sides)

Thus, the answer is 398.

Page No 6.43:

Question 8:

If x-1x=3, find the values of x2+1x2 and x4+1x4.

Answer:

Let us consider the following equation:

x-1x=3

Squaring both sides, we get:

x-1x2=32=9x-1x2=9x2-2×x×1x+1x2=9x2-2+1x2=9
x2+1x2=11                               (Adding 2 to both sides)

Squaring both sides again, we get:

x2+1x22=112=121x2+1x22=121x22+2x21x2+1x22=121x4+2+1x4=121
x4+1x4=119

Page No 6.43:

Question 9:

If x2+1x2=18, find the values of x+1x and x-1x.

Answer:

Let us consider the following expression:

x+1x
Squaring the above expression, we get:

x+1x2=x2+2×x×1x+1x2=x2-2+1x2                       [(a+b)2=a2+b2+2ab]x+1x2=x2+2+1x2
x+1x2=20                                                                     ( x2+1x2=18)
x+1x=±20                                                            (Taking square root of both sides)

Now, let us consider the following expression:

x-1x
Squaring the above expression, we get:

x-1x2=x2-2×x×1x+1x2=x2-2+1x2                     [(a-b)2=a2+b2-2ab]x-1x2=x2-2+1x2
x-1x2=16                                 ( x2+1x2=18)
x-1x=±4                                     (Taking square root of both sides)

Page No 6.43:

Question 10:

If x + y = 4 and xy = 2, find the value of x2 + y2

Answer:

We have:

x+y2=x2+2xy+y2x2+y2=x+y2-2xy
x2+y2=42-2×2                ( x+y=4 and xy=2)
x2+y2=16-4x2+y2=12

Page No 6.43:

Question 11:

If xy = 7 and xy = 9, find the value of x2 + y2

Answer:

We have:

x-y2=x2-2xy+y2x2+y2=x-y2+2xy
x2+y2=72+2×9                     ( x-y=7 and xy=9 )
x2+y2=72+2×9x2+y2=49+18x2+y2=67



Page No 6.44:

Question 12:

If 3x + 5y = 11 and xy = 2, find the value of 9x2 + 25y2

Answer:

We have:

3x+5y2=3x2+23x5y+5y23x+5y2=9x2+30xy+25y29x2+25y2=3x+5y2-30xy
9x2+25y2=112-30×2                          ( 3x+5y=11 and xy=2)
9x2+25y2=121-609x2+25y2=61

Page No 6.44:

Question 13:

Find the values of the following expressions:
(i) 16x2 + 24x + 9, when x=74
(ii) 64x2 + 81y2 + 144xy, when x = 11 and y=43
(iii) 81x2 + 16y2 − 72xy, when x=23 and y=34

Answer:

(i) Let us consider the following expression:

16x2+24x+9

Now

16x2+24x+9=4x+32                                 (Using identity a+b2=a2+2ab+b2)
16x2+24x+9=4×74+32            (Substituting x=74)16x2+24x+9=7+3216x2+24x+9=10216x2+24x+9=100


(ii) Let us consider the following expression:

64x2+81y2+144xy

Now

64x2+81y2+144xy=8x+9y2                                (Using identity a+b2=a2+2ab+b2)
64x2+81y2+144xy=811+9432                  (Substituting x=11 and y=43)64x2+81y2+144xy=88+122 64x2+81y2+144xy=1002 64x2+81y2+144xy=10000

(iii) Let us consider the following expression:

81x2+16y2-72xy

Now

81x2+16y2-72xy=9x-4y2                                  (Using identity a+b2=a2-2ab+b2)
81x2+16y2-72xy=923-4342        (Substituting x=23and y=34)81x2+16y2-72xy=6-32  81x2+16y2-72xy=32 81x2+16y2-72xy=9

Page No 6.44:

Question 14:

If x+1x=9, find the value of x4+1x4.

Answer:

Let us consider the following equation:

x+1x=9

Squaring both sides, we get:

x+1x2=92=81x+1x2=81x2+2×x×1x+1x2=81x2+2+1x2=81
x2+1x2=79                               (Subtracting 2 from both sides)

Now, squaring both sides again, we get:

x2+1x22=792=6241x2+1x22=6241x22+2x21x2+1x22=6241x4+2+1x4=6241
x4+1x4=6239

Page No 6.44:

Question 15:

If x+1x=12, find the value of x-1x.

Answer:

Let us consider the following equation:

x+1x=12

Squaring both sides, we get:

x+1x2=122=144x+1x2=144x2+2×x×1x+1x2=144                      [ (a+b)2=a2+b2+2ab]x2+2+1x2=144
x2+1x2=142                               (Subtracting 2 from both sides)

Now

x-1x2=x2-2×x×1x+1x2=x2-2+1x2                         [(a-b)2=a2+b2-2ab]x-1x2=x2-2+1x2x-1x2=142-2                                                  ( x2+1x2=142)x-1x2=140     x-1x=±140                                                    Taking square root

Page No 6.44:

Question 16:

If 2x + 3y = 14 and 2x − 3y = 2, find the value of xy.
[Hint: Use (2x + 3y)2 − (2x − 3y)2 = 24xy]

Answer:

We will use the identity a+ba-b=a2-b2 to obtain the value of xy.
Squaring (2x+3y) and (2x-3y) both and then subtracting them, we get:
2x+3y2-2x-3y2=2x+3y+2x-3y2x+3y-2x-3y=4x×6y=24xy2x+3y2-2x-3y2=24xy
24xy=2x+3y2-2x-3y224xy=142-2224xy=14+214-2                        ( a+ba-b=a2-b2)24xy=16×12xy=16×1224                                       (Dividing both sides by 24)xy=8

Page No 6.44:

Question 17:

If x2 + y2 = 29 and xy = 2, find the value of
(i) x + y
(ii) x − y
(iii) x4 + y4

Answer:

(i) We have:

x+y2=x2+2xy+y2x+y=±x2+2xy+y2x+y=±29+2×2                (  x2+y2=29 and xy=2)x+y=±29+4x+y=±33

(ii) We have:

x-y2=x2-2xy+y2x-y=±x2-2xy+y2x+y=±29-2×2                 ( x2+y2=29 and xy=2)x+y=±29-4x+y=±25x+y=±5

(iii) We have:

x2+y22=x4+2x2y2+y4x4+y4=x2+y22-2x2y2x4+y4=x2+y22-2xy2x4+y4=292-222                             ( x2+y2=29 and xy=2)x4+y4=841-8x4+y4=833

Page No 6.44:

Question 18:

What must be added to each of the following expressions to make it a whole square?
(i) 4x2 − 12x + 7
(ii) 4x2 − 20x + 20

Answer:

(i) Let us consider the following expression:

4x2-12x+7
The above expression can be written as:

4x2-12x+7=2x2-2×2x×3+7
It is evident that if 2x is considered as the first term and 3 is considered as the second term, 2 is required to be added to the above expression to make it a perfect square. Therefore, 7 must become 9.

Therefore, adding and subtracting 2 in the above expression, we get:

4x2-12x+7+2-2=2x2-2×2x×3+7+2-2=2x2-2×2x×3+9-2=2x+32-2
Thus, the answer is 2.

(ii) Let's consider the following expression:

4x2-20x+20
The above expression can be written as:

4x2-20x+20=2x2-2×2x×5+20
It is evident that if 2x is considered as the first term and 5 is considered as the second term, 5 is required to be added to the above expression to make it a perfect square. Therefore, number 20 must become 25.

Therefore, adding and subtracting 5 in the above expression, we get:

4x2-20x+20+5-5=2x2-2×2x×5+20+5-5=2x2-2×2x×5+25-5=2x+52-5
Thus, the answer is 5.

Page No 6.44:

Question 19:

Simplify:
(i) (x − y)(x + y) (x2 + y2)(x4 + y2)
(ii) (2x − 1)(2x + 1)(4x2 + 1)(16x4 + 1)
(iii) (4m − 8n)2 + (7m + 8n)2
(iv) (2.5p − 1.5q)2 − (1.5p − 2.5q)2
(v) (m2n2m)2 + 2m3n2

Answer:

To simplify, we will proceed as follows:
(i) 
(i) x-yx+yx2+y2x4+y4=x2-y2x2+y2x4+y4            ∵ a+ba-b=a2-b2=x4-y4x4+y4                         ∵ a+ba-b=a2-b2=x8-x8                                       a+ba-b=a2-b2

(ii) 2x-12x+14x2+116x4+1=2x2-124x2+116x4+1             a+ba-b=a2-b2      =4x2-14x2+116x4+1           =4x22-12216x4+1                    a+ba-b=a2-b2=16x4-1  16x4+1        =16x42 - 12                                       a+ba-b=a2-b2=256x8-1

(iii) 7m-8n2+7m+8n2=27m2+28n2                      ∵ a-b2+a+b2=2a2+2b2=98m2+128n2


(iv) 2.5p-1.5q2-1.5p-2.5q2=2.5p2+1.5q2-22.5p1.5q-1.5p2+2.5q2-21.5p2.5q =2.5p2+1.5q2-22.5p1.5q-1.5p2-2.5q2+21.5p2.5q=2.5p2-1.5p2+1.5q2-2.5q2 =2.5p+1.5p2.5p-1.5p+1.5q+2.51.5q-2.5q                   a+ba-b=a2-b2                    =4p×p+4q×-q=4p2-4q2 =4p2-4q2                                                    

v) m2-n2m2+2m3n2=m22+n2m2                                           a-b2+2ab=a2+b2=m4+n4m2 

Page No 6.44:

Question 20:

Show that:
(i) (3x + 7)2 − 84x = (3x − 7)2
(ii) (9a − 5b)2 + 180ab = (9a + 5b)2
(iii) 4m3-3n42+2mn=16m29+9n216
(iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2
(v) (a − b)(a + b) + (b − c)(b + c) + (c − a)( c + a) = 0

Answer:


(i) LHS=3x+72-84x=3x+72-4×3x×7=3x-72          a+b2-4ab=a-b2=RHSBecause LHS is equal to RHS, the given equation is verified.


(ii) LHS=9a-5b2+180ab=9a-5b2+4×9a×5b=9a+5b2                                   a-b2+4ab=a+b2=RHSBecause LHS is equal to RHS, the given equation is verified.


(iii) LHS=4m3-3n42+2mn=4m3-3n42+2×4m3×3n4=4m32+3n42                                 a-b2+2ab=a2+b2=16m29+9n216=RHSBecause LHS is equal to RHS, the given equation is verified.


(iv) LHS=4pq+3q2-4pq-3q2=44pq3q                         a+b2-a+b2=4ab=48pq2 =RHSBecause LHS is equal to RHS, the given equation is verified.


(v) LHS=a-ba+b+b-cb+c+c+ac-a=a2-b2+b2-c2+c2-a2                                                   a+ba-b=a2-b2=a2-b2+b2-c2+c2-a2  =0=RHSBecause LHS is equal to RHS, the given equation is verified.



Page No 6.47:

Question 1:

Find the following products:
(i) (x + 4) (x + 7)
(ii) (x − 11) (x + 4)
(iii) (x + 7) (x − 5)
(iv) (x − 3) ( x − 2)
(v) (y2 − 4) (y2 − 3)
(vi) x+43x+34
(vii) (3x + 5) (3x + 11)
(viii) (2x2 − 3) (2x2 + 5)
(ix) (z2 + 2) (z2 − 3)
(x) (3x − 4y) (2x − 4y)
(xi) (3x2 − 4xy) (3x2 − 3xy)
(xii) x+15(x + 5)
(xiii) z+34z+43
(xiv) (x2 + 4) (x2 + 9)
(xv) (y2 + 12) (y2 + 6)
(xvi) y2+57y2-145
(xvii) (p2 + 16) p2-14

Answer:

(i) Here, we will use the identity x+ax+b=x2+a+bx+ab.
x+4x+7=x2+4+7x+4×7=x2+11x+28

(ii) Here, we will use the identity x-ax+b=x2+b-ax-ab.
x-11x+4=x2+4-11x-11×4=x2-7x-44

(iii) Here, we will use the identity​ x+ax-b=x2+a-bx-ab.
x+7x-5=x2+7-5x-7×5=x2+2x-35

(iv) Here, we will use the identity​ x-ax-b=x2-a+bx+ab.
x-3x-2=x2-3+2x+3×2=x2-5x+6

(v) Here, we will use the identity​ x-ax-b=x2-a+bx+ab.
y2-4y2-3=y22-4+3y2+4×3=y4-7y2+12

(vi) Here, we will use the identity​ x+ax+b=x2+a+bx+ab.
x+43x+34=x2+43+34x+43×34=x2+2512x+1

(vii) Here, we will use the identity​ x+ax+b=x2+a+bx+ab.
3x+53x+11=3x2+5+113x+5×11=9x2+48x+55

(viii) Here, we will use the identity​ x-ax+b=x2+b-ax-ab.
2x2-32x2+5=2x22+5-32x2-3×5=4x4+4x2-15

(ix) Here, we will use the identity​ x+ax-b=x2+a-bx-ab.
z2+2z2-3=z22+2-3z2-2×3=z4-z2-6

(x) Here, we will use the identity​ x-ax-b=x2-a+bx+ab.
3x-4y2x-4y=4y-3x4y-2x                                     (Taking common -1 from both parentheses)=4y2-3x+2x4y+3x×2x=16y2-12xy+8xy+6x2=16y2-20xy+6x2

(xi) Here, we will use the identity​ x-ax-b=x2-a+bx+ab.

3x2-4xy3x2-3xy=3x22-4xy+3xy3x2+4xy×3xy=9x4-12x3y+9x3y+12x2y2=9x4-21x3y+12x2y2

(xii) Here, we will use the identity​ x+ax+b=x2+a+bx+ab.
x+15x+5=x2+15+5x+15×5=x2+265x+1

(xiii) Here, we will use the identity​ x+ax+b=x2+a+bx+ab.
z+34z+43=z2+34+43x+34×43=z2+2512z+1

(xiv) Here, we will use the identity​ x+ax+b=x2+a+bx+ab.
x2+4x2+9=x22+4+9x2+4×9=x4+13x2+36

(xv) Here, we will use the identity​ x+ax+b=x2+a+bx+ab.
y2+12y2+6=y22+12+6y2+12×6=y4+18y2+72

(xvi) Here, we will use the identity​ x+ax-b=x2+a-bx-ab.
y2+57y2-145=y22+57-145y2-57×145=y4-7335y2-2

(xvii) Here, we will use the identity​ x+ax-b=x2+a-bx-ab.
p2+16p2-14=p22+16-14p2-16×14=p4+634p2-4

Page No 6.47:

Question 2:

Evaluate the following:
(i) 102 × 106
(ii) 109 × 107
(iii) 35 × 37
(iv) 53 × 55
(v) 103 × 96
(vi) 34 × 36
(vii) 994 × 1006

Answer:

(i) Here, we will use the identity x+ax+b=x2+a+bx+ab
102×106=100+2100+6=1002+2+6100+2×6=10000+800+12=10812

(ii) Here, we will use the identity x+ax+b=x2+a+bx+ab
109 × 107=100+9100+7=1002+9+7100+9×7=10000+1600+63=11663

(iii) Here, we will use the identity x+ax+b=x2+a+bx+ab
35 × 37=30+530+7=302+5+730+5×7=900+360+35=1295

(iv) Here, we will use the identity x+ax+b=x2+a+bx+ab
53 × 55=50+350+5=502+3+550+3×5=2500+400+15=2915

(v) Here, we will use the identity x+ax-b=x2+a-bx-ab
103 × 96=100+3100-4=1002+3-4100-3×4=10000-100-12=9888

(vi) Here, we will use the identity x+ax+b=x2+a+bx+ab
34 × 36=30+430+6=302+4+630+4×6=900+300+24=1224

(vii) Here, we will use the identity x-ax+b=x2+b-ax-ab
994 × 1006=1000-6×1000+6=10002+6-6×1000-6×6=1000000-36=999964



Page No 6.5:

Question 1:

Add the following algebraic expressions:
(i) 3a2b, − 4a2b, 9a2b
(ii) 23a, 35a, -65a
(iii) 4xy2 − 7x2y, 12x2y − 6xy2, − 3x2y +5xy2
(iv) 32a-54b+25c, 23a-72b+72c, 53a+52b-54c
(v) 112xy+125y+137x, -112y-125x-137xy
(vi) 72x3-12x2+53, 32x3+74x2-x+13, 32x2-52x-2

Answer:

(i) To add the like terms, we proceed as follows:

3a2b+-4a2b+9a2b=3a2b-4a2b+9a2b=3-4+9a2b                    (Distributive law)=8a2b  

(ii) To add the like terms, we proceed as follows:

23a+35a+-65a=23a+35a-65a=23+35-65a                   ( Distributive law)=115a  

(iii) To add, we proceed as follows:

4xy2-7x2y+12x2y+-6xy2+-3x2y+5xy2=4xy2-7x2y+12x2y-6xy2-3x2y+5xy2=4xy2-6xy2+5xy2-7x2y+12x2y-3x2y                ( Collecting like terms )=3xy2+2x2y                                                                (Combining like terms)

(iv) To add, we proceed as follows:

32a-54b+25c+23a-72b+72c+53a+52b-54c=32a-54b+25c+23a-72b+72c+53a+52b-54c=32a+23a+53a-54b-72b+52b+25c+72c-54c                (Collecting like terms)=236a-94b+5320c                                                                        (Combining like terms)

(v) To add, we proceed as follows:

112xy+125y+137x+-112y-125x-137xy=112xy+125y+137x-112y-125x-137xy=112xy-137xy  +125y-112y+137x-125x              (  Collecting like terms)=5114xy-3110y-1935x                                                    ( Combining like terms)

(vi) To add, we proceed as follows:

72x3-12x2+53+32x3+74x2-x+13+32x2-52x-2=72x3-12x2+53+32x3+74x2-x+13+32x2-52x-2=72x3+32x3-12x2+74x2+32x2-x-52x+53+13-2                        ( Collecting like terms)=5x3+114x2-72x                                                                                            ( Combining like terms)

Page No 6.5:

Question 2:

Subtract:
(i) − 5xy from 12xy
(ii) 2a2 from − 7a2
(iii) 2a − b from 3a − 5b
(iv) 2x3 − 4x2 + 3x + 5 from 4x3 + x2 + x + 6
(v) 23y3-27y2-5 from 13y3+57y2+y-2
(vi) 32x-54y-72z from 23x+32y-43z
(vii) x2y-45xy2+43xy from 23x2y+32xy2-13xy
(viii) ab7-353bc+65ac from 35bc-45ac

Answer:

(i) 12xy--5xy=12xy+5xy=17xy

(ii) -7a2-2a2=-7a2-2a2=-9a2

(iii) 3a-5b-2a-b=3a-5b-2a+b=3a-5b-2a+b=3a-2a-5b+b=a-4b

(iv) 4x3+x2+x+6-2x3-4x2+3x+5=4x3+x2+x+6-2x3+4x2-3x-5=4x3-2x3+x2+4x2+x-3x+6-5       (Collecting like terms)=2x3+5x2-2x+1                                  (Combining like terms)


(v) 13y3+57y2+y-2-23y3-27y2-5=13y3+57y2+y-2-23y3+27y2+5=13y3-23y3+57y2+27y2+y-2+5           (Collecting like terms)=-13y3+y2+y+3                                     (Combining like terms)


(vi) 23x+32y-43z-32x-54y-72z=23x+32y-43z-32x+54y+72z=23x-32x+32y+54y-43z+72z            (Collecting like terms )=-56x+114y+136z                                    (Combining like terms)


(vii) 23x2y+32xy2-13xy-x2y-45xy2+43xy=23x2y+32xy2-13xy-x2y+45xy2-43xy=23x2y-x2y+32xy2+45xy2-13xy-43xy         (Collecting like terms)=-13x2y+2310xy2 -53xy                                  (Combining like terms)


(viii) 35bc-45ac-ab7-353bc+65ac=35bc-45ac-ab7+353bc-65ac=35bc+353bc-45ac-65ac-ab7          (Collecting like terms)=18415bc-2ac-ab7                                    (Combining like terms)

Page No 6.5:

Question 3:

Take away:
(i) 65x2-45x3+56+32x from x33-52x2+35x+14
(ii) 5a22+3a32+a3-65 from 13a3-34a2-52
(iii) 74x3+35x2+12x+92 from 72-x3-x25
(iv) y33+73y2+12y+12 from 13-53y2
(v) 23ac-57ab+23bc from 32ab-74ac-56bc

Answer:

(i) The difference is given by:

x33-52x2+35x+14-65x2-45x3+56+32x=x33-52x2+35x+14-65x2+45x3-56-32x
=x33+45x3-52x2-65x2+35x-32x+14-56             (Collecting like terms)
=5+1215x3+-25-1210x2+6-1510x+6-2024
=1715x3-3710x2-910x-712                                            (Combining like terms)

(ii) The difference is given by:

13a3-3a24-52-5a22+3a32+a3-65=13a3-3a24-52-5a22-3a32-a3+65
=13a3-3a32-3a24-5a22-a3-52+65                      (Collecting like terms)
=2-96a3+-3-104a2-a3+-25+1210
=-76a3-134a2-a3-1310                                             (Combining like terms)

(iii) The difference is given by:

72-x3-x25-7x34+3x25+x2+92=72-x3-x25-7x34-3x25-x2-92
=72-92-x3-x2-x25-3x25-7x34                     (Collecting like terms)
=7-92+-2-36x+-1-35x2-7x34
=-1-5x6-4x25-7x34                                           (Combining like terms )

(iv) The difference is given by:

13-53y2-y33+7y23+y2+12=13-53y2-y33-7y23-y2-12
=13-12-y2-53y2-7y23-y33             ( Collecting like terms)
=2-36-y2+-5-73y2-y33
=-16-y2-4y2-y33                               (Combining like terms. )

(v) The difference is given by:

32ab-74ac-56bc-23ac-57ab+23bc=32ab-74ac-56bc-23ac+57ab-23bc

=32ab+57ab-74ac-23ac-56bc-23bc             (Collecting like terms )
=21+1014ab+-21-812ac+-5-46bc
=3114ab-2912ac-32bc                                         (Combining like terms )



Page No 6.6:

Question 4:

Subtract 3x − 4y − 7z from the sum of x − 3y + 2z and − 4x + 9y − 11z.

Answer:

Let first add the expressions x-3y+2z and -4x+9y-11z. We get:

x-3y+2z+-4x+9y-11z    
=x-3y+2z-4x+9y-11z
=x-4x-3y+9y+2z-11z                     (Collecting like terms)              
=-3x+6y-9z                                       (Combining like terms)
                                           
Now, subtracting the expression 3x-4y-7z from the above sum; we get:

-3x+6y-9z-3x-4y-7z=-3x+6y-9z-3x+4y+7z
=-3x-3x+6y+4y-9z+7z          (Collecting like terms)
=-6x+10y-2z                              (Combining like terms)

Thus, the answer is -6x+10y-2z.

Page No 6.6:

Question 5:

Subtract the sum of 3l − 4m − 7n2 and 2l + 3m − 4n2 from the sum of 9l + 2m − 3n2 and − 3l + m + 4n2 .....

Answer:

We have to subtract the sum of (3l - 4m - 7n2) and (2l + 3m - 4n2) from the sum of (9l + 2m - 3n2) and  (-3l m + 4n2)​

9l+2m-3n2+-3l+m+4n2-3l-4m-7n2+2l+3m-4n2
=9l-3l+2m+m-3n2+4n2-3l+2l-4m+3m-7n2-4n2              
=6l+3m+n2-5l-m-11n2                                                 (Combining like terms inside the parentheses)
=6l+3m+n2-5l+m+11n2        
=6l-5l+3m+m+n2+11n2                                                        (Collecting like terms)
=l+4m+12n2                                                                              (Combining like terms)

Thus, the required solution is l+4m+12n2.

Page No 6.6:

Question 6:

Subtract the sum of 2xx2 + 5 and − 4x − 3 + 7x2 from 5.

Answer:

We have to subtract the sum of (2x  x2 + 5) and (4x − 3 + 7x) from 5.

5-2x-x2+5+-4x-3+7x2=5-2x-4x-x2+7x2+5-3=5-2x+4x+x2-7x2-5+3
=5-5+3-2x+4x+x2-7x2                (Collecting like terms)
=3+2x-6x2                                        (Combining like terms)

Thus, the answer is 3+2x-6x2.

Page No 6.6:

Question 7:

Simplify each of the following:
(i) x2 − 3x + 5 − 12 (3x2 − 5x + 7)
(ii) [5 − 3x + 2y − (2xy)] − (3x − 7y + 9)
(iii) 112x2y-94xy2+14xy-114y2x+115yx2+12xy
(iv) 13y2-47y+11-17y-3+2y2-27y-23y2+2
(v) -12a2b2c+13ab2c-14abc2-15cb2a2+16cb2a-17c2ab+18ca2b.

Answer:

(i) x2-3x+5-123x2-5x+7=x2-3x+5-3x22+5x2-72=x2-3x22-3x+5x2+5-72                 (Collecting like terms)=1-32x2+-3+52x+10-72=-x22-x2+32

Thus, the answer is -x22-x2+32.

(ii) 5-3x+2y-2x-y-3x-7y+9=5-3x+2y-2x+y-3x-7y+9=5-5x+3y-3x-7y+9=5-5x+3y-3x+7y-9=5-9-5x-3x+3y+7y=-4-8x+10y

(iii) 112x2y-94xy2+14xy-114y2x+115yx2+12xy
=112x2y+115yx2-94xy2-114y2x+14xy+12xy           (Collecting like terms)
=165+230x2y+-63-228xy2+1+24xy
=16730x2y-6528y2x+32xy                                                (Combining like terms)

(iv) 13y2-47y+11-17y-3+2y2-27y-23y2+2=13y2-47y+11-17y+3-2y2-27y+23y2-2

=13y2-2y2+23y2-47y-17y-27y+11+3-2        (Collecting like terms)
=1-6+23y2+-4-1-27y+12
=-y2-7y+12                                                             (Combining like terms)


(v) -12a2b2c+13ab2c-14abc2-15cb2a2+16cb2a-17c2ab+18ca2b

=-12a2b2c-15cb2a2+13ab2c+16cb2a-14abc2-17c2ab+18ca2b           (Collecting like terms)
=-5-210a2b2c +2+16cb2a2 +-7-428c2ab+18ca2b
=-710a2b2c+12ab2c-1128abc2+18a2bc                                                    (Combining like terms)



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