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Given A = 4x² + y² – 6xy; 

B = 3y² + 12x² + 8xy; 

C = 6x² + 8y² + 6xy 

Write the given expressions in standard form. 

A = 4x² – 6xy + y² 

B = 12x² + 8xy + 3y² 

C = 6x² + 6xy + 8y² 

(i) A + B + C = (4x² – 6xy + y²) + (12x² + 8xy + 3y²) + (6x² + 6xy + 8y²) 

= 4x² – 6xy + y² + 12x² + 8xy + 3y² + 6x² + 6xy + 8y² 

= (4x² + 12x² + 6x² ) + (- 6xy + 8xy + 6xy) + (y² + 3y² + 8y² ) 

= (4 + 12 + 6) x² + (- 6 + 8 + 6) xy + (1 + 3 + 8)y² 

∴ A + B + C = 22x² + 8xy + 12y²

(ii) (A – B) – C 

A + (- B) + (- C) 

Additive inverse of B is 

– B = – (12x² + 8xy + 3y² ) 

∴ – B = – 12x² – 8xy – 3y² 

Additive inverse of C is 

– C = -(6x² + 6xy + 8y² ) 

∴ – C = – 6x² – 6xy – 8y² 

A + (- B) + (- C) 

= (4x² – 6xy + y² ) + (- 12x² – 8xy – 3y² ) + (- 6x² – 6xy – 8y²) 

= 4x² – 6xy + y² – 12x² – 8xy – 3y² – 6x² – 6xy – 8y² 

= (4²x – 12x² – 6x² ) + (- 6xy – 8xy – 6xy) + (y² – 3y² – 8y² )

∴ (A – B) – C = – 14x² – 20xy – 10y²

Let the subtracted algebraic expression may be B’ say

3x² – 4y² + 5xy – B = – x² – y² + 6xy + 20

B = (3x² – 4y² + 5xy) – (- x² – y² + 6xy + 20)

= 3x² – 4y² + 5xy + x² +y² – 6xy – 20

= (3x² + x²) ( – 4y² + y²) + (5xy – 6xy) – 20

∴ B = 4x² – 3y² – xy – 20

∴ The required subtracted expression is B 

= 4x² – 3y² – xy – 20

Like terms are (2x, – 6x) (5y², 18y²).

Given 4x²-7xy+4y2-3, 5+6y2-8xy+x² and 6-2xy+2x²-5y2

To add the given expression we have arrange them column wise is given below:

4x²-7xy+4y2-3

x²-8xy+6y2+5

2x²-2xy-5y2+6

(i) 7xy – 2xy = (7 – 2) xy = 5xy

(ii) 10a² – 4a² = (10 – 4)a² = 6a²

(iii) 3p – 15p = (3 – 15)p = – 12p

(iv) – 20 m² n – 6m² n = (- 20 – 6) m² n = – 26m² n

(v) – a² b² – a² b² = (- 1 – 1)a² b² = – 2a² b²

\(\frac{-1}{27}\times \frac{9}{2}\times a^2\times a^3\times b^2\times b^2\times c^2\)

= \(\frac{-1}{6}\times a^5\times b^4\times c^2\)

= \(\frac{-1}{6}a^5b^4c^2\)

i) 2xy, 7xy

7xy – 2xy = (7 – 2) xy = 5xy

ii) 5a², 10a²

10a -5a = (10-5) a² = 5a²

iii)  12y, 3y

3y – 12y = (3 – 12)y = -9y

iv) 6x²y, 4x²y

4x²y – 6x²y = (4 – 6) x²y = -2x²y

v) 6xy,-12xy

(-12xy) – 6xy = (-12 – 6) xy = -18xy

–5a²b and –5b²a are unlike terms.

The terms having different algebraic factors are called unlike terms.

Given 3x – (y – 2x)

Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them. Therefore, we have

= 3x – y + 2x

On simplifying, we get

= 5x – y

(-4) × (-6) – (-3) × x² × x × y² × y × z²

= - 72 × x³ × y³ × z²

= -72x³y³z²

Correct option is  A) x² y² z² 

Correct option is (A) x² y² z²

P = xy, Q = yz, R = zx.

Then PQR = (xy) (yz) (zx)

\(=(x\times x)(y\times y)(z\times z)\)

= \(x^2y^2z^2\).

ABC = xy × yz × zx = x²y²z²

(c) 4 

  \(\frac{x^2 + y^2 + z^2 - 64}{xy - yz-zx}\)= –2 

⇒ x² + y² + z² = - 2xy + 2yz + 2zx + 64

Now,

(x + y + z)² =  x² + y² + z² + 2xy + 2yz + 2zx

⇒ (3z + z)² = - 2xy + 2yz + 2zx + 64 + 2xy + 2yz + 2zx

⇒ 16z² = 4yz + 4xz +64

⇒ 16z² = 4z(x + y) + 64 ⇒ 16z² = 4z (3z) + 64

⇒ 16z² - 12z² = 64

⇒ 4z² = 64 ⇒ z² = 16 ⇒ z = 4.

(x + y + z)(x² + y² + z² - xy - yz - zx) = x³ + y³ +z³ - 3xyz

Given, x + y + z = 1, xy +yz + zx = -1 and xyz = -1

To find, x² + y² + z², we use the identity :

(x + y + z)² = x² + y² + z²  + 2( xy + yz + zx)

⇒ 1 =  x² + y² + z² + (2 x -1)

⇒  x² + y² + z² = 1 + 2 = 3

∴ Substituting all the values in eqn (i), we get

1 x (3 - (-1)) =  x³ + y³ + z³ - (3 x -1)

⇒ 1 = (3 + 1) = x³ + y³ + z³ +3

⇒ 4 = x³ + y³ + z³ + 3 ⇒ x³ + y³ + z³ = 1.

-7 ×\(\frac{1}{4}\) × x × y × x² × y × z

= \(\frac{-7}{4}\)× x³ × y² × z

= \(\frac{-7}{4}\)x³y²z

(-5) × (-10) × (-2) × a × a² × a³

= -100 × a⁶

= -100a⁶

(i) Additive inverse of – 6a = – (- 6a) = 6a

(ii) Standard form of 2 + 7c² = 7c² + 2 

Additive inverse of 7c² + 2 

= – (7c² + 2) 

= – 7c² – 2

(iii) Given expression is in standard form. Additive inverse of 6x² + 4x – 5 

= – (6x² + 4x – 5) 

= – 6x² – 4x + 5

(iv) Standard form of 3c + 7a – 9b = 7a – 9b + 3c 

Additive inverse of 7a – 9b + 3c 

= – (7a – 9b + 3c) 

= – 7a + 9b – 3c

Let A = 6a – 2b + 3c, B = 3a + 4b – 2c 

Subtracting 3a + 4b – 2c from 6a – 2b + 3c is equal to adding additive inverse of 3a + 4b – 2c to 6a – 2b + 3c 

i.e., A – B = A + (-B) 

additive inverse of (3a + 4b – 2c) = – (3a + 4b – 2c) 

= – 3a – 4b + 2c A – B = A + (-B) 

= 6a – 2b + 3c + (- 3a – 4b + 2c) 

= 6a – 2b + 3c – 3a – 4b +2c 

= (6 – 3)a – (2 + 4)b + (3 + 2)c 

Thus, the required answer = 3a – 6b + 5c

6a² + 3ab + 5b² – 2ab – b² + 2a² + 4ab + 2b² – a²

= (6a² + 2a² – a² ) + (3ab – 2ab + 4ab) + (5b² – b² + 2b² ) 

= [(6 + 2 – 1) a² ] + [(3 – 2 + 4)ab] + [(5 – 1 + 2)b² ] 

= 7a² + 5ab + 6b²

7 × -5 × 6 × a × a × a × b × b² × b × c × c²

= 210 × a³ × b⁴ × c³

= 210a³b⁴c³

(i) False

(ii) False

(iii) True

(iv) False

(v) True.

Given cost of each pen is ₹ x and cost of each pencil is ₹ y. 

Arjun bought 3 pens and 2 pencils. 

Cost of 3 pens = 3 × ₹ x = ₹ 3x 

Cost of 2 pencils = 2 × ₹y = ₹ 2y 

Amount paid by Arjun = 3x + 2y = ₹ (3x + 2v) 

George bought one pen and 4 pencils. 

Cost of 1 pen = 1 × ₹ x = ₹ x 

Cost of 4 pencils = 4 × ₹ y = ₹ 4y 

Amount paid by George = x + 4y = ₹(x + 4y) 

Total bill amount = Arjun amount + George amount 

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

 3x + 2y + x + 4y 

= 3x + x + 2y + 4y 

= (3 + 1)x + (2 +4)y 

∴ Total bill amount = ₹ (4x + 6y)

Given terms are 3a, 6b, 5c, – 8a, 7c, 9c, – a, 2/3 b, 7c/9, a/2

Like terms: 3a, – 8a, – a, a/2

6b, 2/3 b

5c, 7c, 9c, 7c/9

Correct option is  B) a – b

Correct option is  C) \(\frac{a^2}{9}-\frac{b^2}{25} \)

Correct option is (C) \(\frac{a^2}9-\frac{b^2}{25}\)

\((\frac a3+\frac b5)(\frac a3-\frac b5)=(\frac a3)^2-(\frac b5)^2\) \(=\frac{a^2}9-\frac{b^2}{25}\)

 Correct option is  B) 3

Correct option is   D) None

Correct option is (D) None

\((a–b)(a^2+ab+b^2)\) \(=a^3+a^2b+ab^2-a^2b-ab^2-b^3\)

\(=a^3-b^3\)

A) x² + 5x + 6

(i) (x +1)(x – 1)(x² + 1) 

We know that, from formula, 

(a + b)(a – b) = a² – b² 

(x + 1)(x – 1) (x² + 1) = (x² – 1)(x² + 1) 

= (x²)² – 1 = x⁴ – 1

(ii) (x- 3)(x + 3)(x² + 9) 

We know that, from formula, 

(a + b)(a – b) = a² – b² 

(x – 3)(x + 3)(x² + 9) 

= (x² – 9)(x² + 9) 

= (x²)² – 9² = x⁴ – 81

(iii) (3x – 2y)(3x + 2y)(9x² + 4y²) 

We know that, from formula, 

(a + b)(a – b) = a² – b² 

(3x – 2y)(3x + 2y)(9x² + 4y²) 

= (9x² – 4y²)(9x² + 4y²) 

= 81x⁴ – 16y⁴

(iv) (2p + 3)(2p – 3)(4p² + 9) 

We know that, from formula, 

(a + b)(a – b) = a² – b² 

(2p + 3)(2p – 3)(4p² + 9) 

= (4p² – 9)(4p²+ 9) 

= (4p²)² – 9² = 16p⁴ – 81

(i) 101² – 99²

= (101 + 99) (101 – 99) (Using given identity)

= (200) (2)

= 400

(ii) (10.3)² – (9.7)²

= (10.3 + 9.7) (10.3 – 9.7) (Using given identity)

= (20) (0.6)

= 12

(iii) 153² – 147²

= (153 + 147) (153 – 147) (Using given identity)

= (300) (6)

= 1800

Correct option is  C) 99.96

Correct option is (C) 99.96

9.8 \(\times\) 10.2 = (10 - 0.2) \(\times\) (10+0.2)

\(=10^2-(0.2)^2\)

= 100 - 0.04 = 99.96

Correct option is  D) 9801

A) (3x + y) (3x – y)

Given x³ + 2x²y + 6xy² − y³ and y³−3xy²−4x²y

= (y³ – 3xy² – 4x²y) – (x³ + 2x²y + 6xy² – y³)

= y³ – 3xy² – 4x²y – x³ – 2x²y – 6xy² + y³

= y³ + y³– 3xy²– 6xy²– 4x²y – 2x²y – x³

= 2y³– 9xy² – 6x²y – x³

-17x²(3x – 4)

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P × (q – r) = (p × q) – (p × r)

Now,

= (-17x² × 3x) – (-17x² × 4)

= (-51x³ + 68x²) … [∵ a^m × aⁿ = a^m+n]

(7/2)x²((4/7)x + 2)

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P × (q + r) = (p × q) + (p × r)

Now,

= ((7/2)x² × (4/7)x) + ((7/2)x² × 2)

= ((7×4)/ (2×7))x³ + ((7×2)/(2×1))x² … [∵ a^m × aⁿ = a^m+n]

= ((1×2)/ (1×1))x³ + ((7×1)/(1×1))x²

= (2)x³ + (7)x²

-4x²y(3x² – 5y)

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P × (q – r) = (p × q) – (p × r)

Now,

= (-4x²y × 3x²) – (-4x²y × -5y)

= (-12x⁴y + 20x²y²) … [∵ a^m × aⁿ = a^m+n]

(x³ – y³) (x² + y²)

Suppose (a – b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) × (c + d) = a × (c + d) – b × (c + d) = (a × c + a × d) – (b × c + b × d)

= ac + ad – bc – bd

Let,

a= x³, b= y³, c= x², d= y²

Now,

= x³ × (x² + y²) – y³ × (x² + y²)

= [(x³ × x²) + (x³ × y²)] – [(y³ × x²) + (y³ × y²)]

= [x⁵ + x³y² – y³x² – y⁵]

Correct option is  D) p³ + q³

Correct option is (D) p³ + q³

\((p+q)(p^2-pq+q^2)\) \(=p^3-p^2q+pq^2+p^2q-pq^2+q^3\)

= \(p^3+q^3\).

(a – b)² = a² – 2ab + b² 

Here a = x, b = 6 

(x – 6)² = x² – (2)(x)(6) + 6² 

= x² – 12x + 36

((3/4)a + (2/3)b) (4a + 3b)

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) × (c + d) = a × (c + d) + b × (c + d) = (a × c + a × d) + (b × c + b × d)

= ac + ad + bc + bd

Let,

a= (3/4)a, b=(2/3)b, c= 4a, d= 3b

Now,

= (3/4)a × (4a + 3b) + (2/3)b × (4a + 3b)

= [((3/4)a × 4a) + ((3/4)a + 3b)] – [((2/3)b × 4a) + ((2/3)b × + 3b)]

= [3a² + (9/4)ab + (8/3)ab + 2b²]

= [3a² + ((27+32)/12)ab + 2b²]

= [3a² + (59/12)ab + 2b²]

A) (m² – n² ) (m + n)

41² = (40 + 1)² (a + b)² = a² + 2ab + b² 

Here a = 40, b = 1 

(40 + 1)² = 40² + 2 × 40 × 1 + (1)² 

= 1600 + 80 + 1 

41² = 1681

53² = (50 + 3)² (a + b)² 

= a² +2ab + b² 

Here a = 50, b = 3 

(50 + 3)² = 50² + 2 × 50 × 3 + (3)² 

= 2500 + 300 + 9 

53² = 2809

(a – b)² = a² – 2ab + b² 

Here a = 3x, b = 5y 

(3x – 5y)² = (3x)² - (2)(3x)(5y) + (5y)² 

= 9x² – 30xy + 25y²

(a – b)² = a² – 2ab + b² 

Here a = 5a, b = 4b 

(5a – 4b)² = (5a)² – (2)(5a)(4b) + (4b)² 

= 25a² – 40ab + 16b²

 (x² – a²) (x – a)

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) × (c – d) = a × (c – d) – b × (c – d) = (a × c – a × d) – (b × c – b × d)

= ac – ad – bc + bd

Let,

a= x², b= a², c= x, d= a

Now,

= x² × (x – a) – a² × (x – a)

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

= [x³ – x²a – a²x + a³]

(2x² – 5y²) (x² + 3y²)

Suppose (a – b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) × (c + d) = a × (c + d) – b × (c + d) = (a × c + a × d) – (b × c + b × d)

= ac + ad – bc – bd

Let,

a= 2x², b= 5y², c= x², d= 3y²

Now,

= 2x² × (x² + 3y²) – 5y² × (x² + 3y²)

= [(2x² × x²) + (2x² × 3y²)] – [(5y² × x²) + (5y² × 3y²)]

= [2x⁴ + 6x²y² – 5y²x² – 15y⁴]

= [2x⁴ + x²y² – 15y⁴]

i) 92² – 8² is in the form of 

a² – b² = (a + b) (a – b).

92² – 8² = (92 + 8)(92 – 8)

= 100 × 84

= 8400

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

= (1000) (968) [∵ (a + b)(a – b) = a² – b²]

= 9,68,000

Correct option is  A) 10p³

Correct option is (A) 10p³

\(–5\,p^2\times-2p\) \(=(–5\times-2)\times(p^2\times p)\)

= \(10p^3\).