WBBSE Solutions For Class 7 Maths Algebra Chapter 4 Algebraic Operations Exercise 4 Solved Problems

WB Class 7 Math Solution Algebra Chapter 4 Algebraic Operations Exercise 4 Solved Problems

1. The term 5, 6, \(\frac{3}{4}\) etc are called constant term and are donated by a, b, c, etc. The term which is always changing are known as variable and are donated by x, y, z etc.

2. The term (7x – 2) and 9x are called algebraic expressing.

3. Relations formed by addition, subtraction, multiplication, and division of few variables and few constants are called an algebraic expression.

4. In expression (5p + 4), p is variable and 5 and 4 are constant.

5. In the algebraic expression 7y, the variable y is multiplied by the constant 7. The factors of 7y are 1, 7, y, and 7y.

6. 7y has only one term and is called a Monomial.
(5p+ 4) has two terms and it is called Binomials.

Read and Learn More WBBSE Solutions for Class 7 Maths

Like terms and Unlike terms:

⇒ If the terms of algebraic expression are alike. They are called like terms and if they are not alike, they are called, unlike terms.

⇒ 3a²b and 4a²b  are like terms. But 3a²b and 4a²b are unlike terms.

⇒ Coefficients: A coefficient is a numerical factor present with any term.
⇒ In 5x², the coefficient of x² is 5 and the coefficient of 5x is x.

Wbbse Class 7 Maths Solutions

Question 1. Choose the correct answer 

1. The sum of (-5x + 3y) and (18x – 15y) is

1. 23x 12y
2. 13x 12y
3. 13x 18y
4. None of these

Solution: -5x + 3y+ 18x-15y
=-5x+18x + 3y 15y
= 13x 12y

So the correct answer is

2. The value of (- 3m² + 2m+ 2) – (m² – 2) is

1. -2m² + 2m
2. -4m² + 2m + 4
3. -2m²+ 2m – 4
4. -2m² + 2m + 4

Solution: (- 3m² + 2m+ 2) – (m² – 2)

= -3m² + 2m + 2-m² + 2
= -3m² -m² + 2m + 2 + 2
=-4m² + 2m + 4

So the correct answer is 2. -4m² + 2m + 4

3. The value of (-3a²) x (4a²b) x (-2) is

1.  24a²b
2. 24a²b³
3. -24a⁴b
4. 24a⁴b

Solution: (-3a²) x (4a²b) x (-2)
=(-3) x (4) x (-2) x a²⁺² x b
= 24a⁴b

So the correct answer is 4. 24a⁴b

Question 2. Write ‘true’ or ‘false’

1. The number of factors of x²y is 6

Solution:

 

The factors of x²y is 1, x, x², y, xy, and i.e., the number of factors is 6.

So the statement is true.

2. The coefficient of x² in (5-6x²y² + 6xy) is 6y

Solution: In the expression (5-6x²y² + 6xy) the term with x is (-6x² y²) whose coefficient of x² is -6y²

So the statement is false.

Wbbse Class 7 Maths Solutions

3. The value of (-48x⁹ + 12x⁶) + 3x³ is (-16x⁶+ 4x³)

Solution: \(\frac{-48 \dot{x}^9+12 x^6}{3 x^3}\)

= \(-\frac{48 x^9}{3 x^3}+\frac{12 x^6}{3 x^3}\)

=-16x^9-3 +4x^6-3 =-16x⁶+4x³

So the statement is true.

Question 3. Fill in the blanks

1. 4x, (5x-2), (7x + 3) together are called _____

Solution: Algebrain expression

2. The product of a²b and (3a-4b) is _____

Solution: a²b (3a – 4b) = 3a³b – 4a²b²

3. If the price of x dozen of pen is ₹ (xy²-7x) then the price of 4 such pen is ₹ _____

Solution: x dozen = 12x

⇒ The price of the number of 12x pen is ₹ (xy² – 7x)

⇒ The price of 1 such pen is ₹ \(\frac{x y^2-7 x}{12 x}\)

⇒ The price of 4 such pen is ₹  \(\frac{4\left(x y^2-7 x\right)}{12 x}\)

=

= ₹ \(\left(\frac{y^2-7}{3}\right)\)

Some important rules: a is non zero integers and also m and n are integers

1. a^m. aⁿ = a^{m + n}

2. a^m ÷ aⁿ = a^{m – n}

3. (a^m)ⁿ = a^mn

4. \(a^m=\frac{1}{a^{-m}}\)

5. a⁰ = 1

6. (ab)^m = a^m.bⁿ

7. \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)

Proof: 

1. \(a^0=a^{m-m}=\frac{a^m}{a^m}=1\)

2. \(a^m=a^{0-(-m)}=\frac{a^0}{a^{-m}}=\frac{1}{a^{-m}}\)

Wbbse Class 7 Maths Solutions

Question 4. Represent the following algebraic expressions into ‘factor free’ type of figure mentioning the prime factors of each term. Also mention the types these expressions with respect to their number of terms.

1. xy + yz + zx
2. x² + x + 1

Solution:

1.

 

2.

 

Question 5. Find the numerical coefficients of the terms, other than the constant term.

1. x² + 6x + 5
2. x – 7xy + 9y

Solution: 1. x² + 6x + 5
The numerical co-efficient of x² is 1.
The numerical co-efficient of x is 6.

2. x – 7xy + 9y

The numerical co-efficient of x, xy, and y are 1, -7, and 9 respectively.

Question 6. Add

1. 7a²b+3ab²-9, 4a²b – 5ab² + 10, a²b + ab² – 15
2. xy-3yz.+4zx, 3xy+5yz-6zx, -2xy + 3yz-zx

Solution:

1. 7a²b+3ab²-9, 4a²b – 5ab² + 10, a²b + ab² – 15

= 7a²b+ 4a²b + a²b+3ab² – 5ab² +ab² – 9+ 10 – 15
= 12a²b – ab² – 14

Another method:

7a²b+3ab²-9
4a²b-5ab² + 10
+ a²b + ab² – 15

= 12a²b- ab² – 14

2. (xy 3yz+4zx) + (3xy + 5yz-6zx) + (-2xy + 3yz – zx)

=
= xy + 3xy – 2xy – 3y + 5yz + 3yz + 4zx-6zxzx
= 2xy + 5yz-3zx.

Question 7. Subtract

1. (3x² – 4xy- 5y²) from (-7x²-3xy+4y²)
2. (m³ – 3m² + 4m – 5) from (m³+ 3m² – 6m + 2)

Solution:

1. (-7x² -3xy + 4y²) – (3x²-4xy – 5y²)

= -7x²-3xy+4y²-3x² + 4xy + 5y²
= -7x²-3x² – 3xy + 4xy+4y²+5y²
= -10x² + xy + 9y²

Alternative method

2. (m³+ 3m²– 6m+ 2) – (m³-3m² + 4m -5)
= m³+3m²-6m+2 -m³ + 3m² – 4m + 5

= 6m2 10m +7

Question 8. How much must be added to (6y² – 6y+ 1) to get (-15y²+ 7y – 7)

Solution:

The required expression is (-15y² + 7y – 7) (6y² – 6y+1)
=-15y²+ 7y-7-6y²+6y+ 1
=-15y²+ 13y-6

Question 9. What must be subtracted from (9x²-3x+7) to get (-3x²+6x-13)

Solution:

The required expression is (9x²-3x+7)-(-3x²+6x-13)
= 9x² – 3x + 7 + 3x²-6x + 13
= 12x² – 9x+20

Wbbse Class 7 Maths Solutions

Question 10. Subtract the sum of (-5y² + 3y -7) and (9y²-7y+ 12) from the sum of (y²+2y-6) and (12y – y² + 5)

Solution:

The required expression is
{(y²+2y-6)+(12y-y²+ 5)} – {(-5y2+ 3y 7) + (9y2-7y+ 12)}

=

= (14y -1)- (2y²-4y+5)
=14y -1 – 2y² + 4y – 5
=-2y²+18y-6

Question 11. Multiply

1. \(\left(-\frac{3}{5} a^2 b\right) \times\left(\frac{15}{8} a b^2 c\right) \times\left(\frac{4}{9} b c^2\right)\)

2. \(\left(\frac{5}{3} a^2 b c\right) \times\left(-\frac{4}{5} a b^2 c\right) \times\left(-\frac{6}{15} a b c^2\right)\)

3. (3a-4b)(5a + 6b)

4. (a² – b² + c²)(a² + b² – c²)

Solution:

1. \(\left(-\frac{3}{5} a^2 b\right) \times\left(\frac{15}{8} a b^2 c\right) \times\left(\frac{4}{9} b c^2\right)\)

2. \(\left(\frac{5}{3} a^2 b c\right) \times\left(-\frac{4}{5} a b^2 c\right) \times\left(-\frac{6}{15} a b c^2\right)\)

3. (3a-4b)(5a + 6b)

= 3a (5a + 6b) – 4b (5a + 6b)
= 15a² + 18ab – 20ab – 24b²
=15a²-2ab-24b²

4. (a²-b²+c²)(a²+b²-c²)

Wbbse Class 7 Maths Solutions

Question 13. Divide the first expression by the second.

1. 20a²b-25ab³c²-30a²bc³, 5ab
2. 6abc-18a²bc² + 24a³b²c³, -6abc
3. a²b⁴ + a4b³ – a³b⁵4, -a⁴b

Solution:

1.

2.

 

3.

 

Question 14. If A = 2x + 3y – 4z, B= 2y+ 3z4x and C2z+3x4y. Find the sum of (A+B+C) and (A B+ C)

Solution:

Given

If A = 2x + 3y – 4z, B= 2y+ 3z4x and C2z+3x4y.

(A+B+ C) + (A- B+ C)
= 2A + 2C
= 2(2x + 3y 4z) + 2(2z+3x-4y)
=4x+6y8z+4z+6x-8y
= 10x 2y4z.

Question 15. If P = 3x² + 2xy + y², Q=-3x² – 2xy + y² and R = x² + y² and x = y – 2. Find the value of (P+Q + R).

Solution:

Given

If P = 3x² + 2xy + y², Q=-3x² – 2xy + y² and R = x² + y² and x = y – 2.

= x² + 3y²
=(-2)² + 3 (-2)² [Putting the value of x and y]
= 4+3×4
= 16.

Question 16. Simplify the following

1. 3x²+ 2x(x + 3) -4x (3x-5)
2. (a² + b²)(a² – b²) + (b² + c²)(b² – c²) + (c² + a²)(c² – a²)
3. (x²+5x)(x -1) – (x² + 2x)(x 1) – 5x (x + 1)
4. (x+3)(x-3)+(x+4)(x-4) – 2x² + 25
5. (x² + 2x – 1)(x − 1) + (x² – 2x + 1)(x + 1) − 2x(x − 1)

Solution:

1. 3x²+ 2x(x + 3) -4x (3x-5)

= 3x²+2x²+6x-12x² + 20x
= −7x² + 26x

2. (a² + b²)(a² – b²) + (b² + c²)(b² – c²) + (c² + a²)(c² – a²)

= a²(a²-b²) + b²(a² – b²) + b²(b² – c²) + c²(b² – c²) + c² (c² – a²) + a² (c² – a²)

= 0

3. (x² + 5x)(x – 1) – (x² + 2x)(x – 1) – 5x (x + 1)

= x²(x -1) + 5x(x – 1) – x² (x – 1) – 2x(x – 1)- 5x(x + 1)

= -2x² – 8x

4. (x+3)(x-3)+(x+4)(x-4)- 2x² + 25.

= x(x-3)+3(x-3) + x(x-4)+4(x-4) – 2x² + 25

= 0

5. (x² + 2x-1)(x 1) + (x² – 2x + 1)(x + 1) 2x(x-1)

= x²(x-1) + 2x(x − 1) − 1(x − 1) + x² (x + 1) − 2x(x + 1) + 1(x + 1) − 2x(x-1)

= 2x³ – 2x² – 2x + 2

 

WB Class 7 Math Solution Algebraic Operations

Algebraic Operations Exercise 6.1

Let’s write the algebraic expression and find the number of terms.

4x, 3x + 1, 2x + 1, 6p – 1, 3y + 6

Class 7 Math Solution WBBSE Algebraic Operations Exercise 6.2

Question 1. Let’s draw ‘factor tree’ type figures and from there let’s find the number of terms and factors of the following algebraic expression :

1. 2x + 1
Solution:

2. 3y+6
Solution:

Question 2.  Let’s study the algebraic expressions given below and fill in the gaps accordingly.

Class 7 Math Solution WBBSE Algebraic Operations Exercise 6.3

Question 1. Let’s form algebraic expressions from the statements given below-

1. y is added to x
Solution: x + y

2. x is subtracted from z
Solution: z – x

3. q is added to twice of p
Solution: 2p + q

4. Multiply x with a square of x.
Solution: x².x = x³

5. \(\frac{1}{4}\) th of the sum of x and y
Solution: \(\frac{x+y}{4}\)

6. 7 is added to 4 times the product of a & b
Solution:  4ab + 7

7. Half of y is added to twice of x
Solution: 2x +\(\frac{1}{2}\) y

8. The product of x and y is subtracted from sum of x and y.
Solution: (x + y) – xy.

Question 2. Let’s Let’s observe the patterns match and fill in the chart

1. 

General Expression 5x + 2 = Number of Match Sticks.

2. Let’s form the general expression with a variable

General Expression, Number of sticks = 4x + 1 where x = No. of trapezium

Question 3. Let us represent the following algebraic expressions into ‘factor tree’ type of figure mentioning the prime factor of each term. Also mention thetypes of these expressions with respect to their number to terms.

1. 5x (Monomial)
Solution:

2. 7 + 2x + x² (Trinomial)
Solution:

3. x² + x + 1 (Trinomial)
Solution:

4. 2x²y + 7 (Binomial)
Solution:

5. 2y³ + y (Binomial)
Solution:

6. x²y+xy²+xyz
Solution:

7. x²y + 2x²y² (Binomial)
Solution:

8. 5x + 2y (Binomial)
Solution:

Question 4. Let’s find the numerical coefficients of the terms, other than constant
term:

1. 2x + 3y
Solution: 2, 3.

2. x² + 2x + 5
Solution: 1, 2.

3. x + 5xy – 7y
Solution: 1, 5, – 7.

4. – 5 – z
Solution: – 1

5. x³ + x – y 
Solution: – 1,1, 1,

6. \(\frac{x}{2}+4\)
Solution: \(\frac{1}{2}\)

Question 5. In the following algebraic expression let’s find the coefficient of x in the terms or terms which have ‘x’ as their factor.

1. y³x + y²
Solution: y³x, y³

2. 5z² – 8zx
Solution: – 8zx, – 8z

3. – x – y + 2
Solution:– x, – 1

4. 4 + y +.yx
Solution: yx, y

5. 2 + x + xy²
Solution: x, 1 , xy2, y2

6.  5xy⁴-1 4
Solution: 15xy⁴, I5y⁴

Question 6. Let’s group the like terms from the algebraic expressions given below. 2x, y, 12xy, 13y², – 5x, 18y, – 4xy, – 2y², 21x²y, 3x, 3xy, – xy, – y, – 6x², -1 5x².
Solution:

2x, y, 12xy, 13y², – 5x, 18y, – 4xy, – 2y², 21x²y, 3x, 3xy, – xy, – y, – 6x², -1 5x²=

(2x, – 5x, 3x); (y, 18y, – y), (12xy, – 4xy, 3xy, – xy), (13y², – 2y²), (21x²y), (- 6x², – 15x²)

Question 7. Let’s identify the like and unlike pairs of terms with reasons from the pairs of terms given below.

1. 2x, 3y
2. 7x, 8x
3. 29x, 6Tx
4. 4xy, 6yz
5. 15 yx, 8xy
6. 5xy, 6x²y²

Solution:

Like term: 2,3,5

Unlike term: 1,4,5

Question 8. From the algebraic expressions given below, let’s write the terms which contain x². And also find the coefficient of x².

1. 5- xy²
Solution: No

2. – 6x² – 8y
Solution: -6

3. 3x² – 1 5xy² – 8y²
Solution: 3

4. 2 + 3k²y+4x
Solution: + 3y

5. 5 – 6x²y² + 6xy
Solution: – 6y²

Class Vii Math Solution WBBSE Algebraic Operations Exercise 6.4

Question 1. Let’s add the following

1. (- 5x + 3y) and (1 8x – 1 5y)
Solution :

Given

(- 5x + 3y) and (1 8x – 1 5y)

⇒ (- 5x + 3y) + (1 8x – 1 5y)

= – 5x + 1 8x + 3y – 1 5y

= 1 3x – 1 2y

2. (7a – 8b + 2c) and (2a + 3b – d)
Solution :

Given

(7a – 8b + 2c) and (2a + 3b – d)

⇒ (7a – 8b + 2c) + (2a + 3b – d)

= 7a + 2a + (- 8b + 3b) + 2c – d

= 9a – 5b + 2c – d

Question 2. Let’s subtract:

1.  (-mn – m + n) from (4mn + m + n)
Solution :

Given

(-mn – m + n) and (4mn + m + n)

⇒ (-mn – m + n) – (4mn + m + n)= (4mn + m + n) – (-mn – m + n)

= 4mn + m + n + mn + m- n

= 4mn + mn + m + m + n- n

= 5mn + 2m

2. (2q² + 3P² – qp + pq²) from (p² + q² – pq + p²q)
Solution:

Given

(2q² + 3P² – qp + pq²) and (p² + q² – pq + p²q)

= (p² + q² – pq + p²q) – (2q² + 3p² – pq + pq²)

= p² + q² – pq + p²q – 2q² – 3p² + pq – pq²

= p² – 3r² + q² – 2q² – pq + pq + p²q – pq²

= – 2p² – q² + p²q – pq²

Class Vii Math Solution WBBSE Algebraic Operations Exercise 6.5

Question 1. Lefs add the algebraic expressions (2x² + x + 2) and (x² + 2x + 2) with the use of colored cards.
Solution:

Let’s add :

Given

(2x² + x + 2) and (x² + 2x + 2)

⇒ (2x² + x + 2) + (x² + 2x + 2)

= 2x² + x² x + 2x + 2 + 2

= 3x² + 3x + 4

Question 2. Let’s subtract the algebraic expressions (3x² + 3x – 2) from (5x² – 2x – 3)
Solution:

Given

(3x² + 3x – 2)and (5x² – 2x – 3)

Let’s subtract the algebraic expressions

(3x² + 3x – 2) from (5x² – 2x – 3)

= (5x² – 2x – 3) – (3x² + 3x – 2)

= 5x² – 2x – 3 – 3x² – 3x + 2

= 5x² – 3x² – 2x – 3x – 3 + 2

= 2x² – 5x 1

Algebraic Operations Exercise 6.6

Question 1. Let’s add the following :

1. (- 5x + 3y) and (18x – 1 5y)
Solution:

Given

(- 5x + 3y) and (18x – 1 5y)

– 5x + 3y + 18x – 15y = – 5x + 18x + 3y – 15y

= 13x – 1 2y

2. (7a – 8b + 2c) and (2a + 3b – d)
Solution:

Given

(7a – 8b + 2c) and (2a + 3b – d)

7a – 8b + 2c + 2a + 3b – d = 7a + 2a – 8b + 3b + 2c – d

= 9a – 5b + 2c> d

Question 2. Let’s Let’s subtract.:

1. (- mn – m + n) from (4mn + m + n)
Solution:

Given

(- mn – m + n) and (4mn + m + n)

(4mn + m + n). – (- mn – m + n)

= 4mn + m + n + mn + m- n

= 4mn + mn + m + m + n- n

= 5mn + 2m

2. (2q² + 3p² – qp + pq²) from (p² + q² – pq + p2q)
Solution:

Given

(2q² + 3p² – qp + pq²) and (p² + q² – pq + p2q)

(p² + q² – pq + p²q) – (2q² + 3p²– qp + pq²)

p² + q² – pq + p²q – 2q² – 3p² + pq – pq²

= p² – 3p² + q² – 2q² – pq + pq + p²q – pq²

= – 2p² – q² + p²q – pq²

= p²q – 2p² – q² – pq²

Algebraic Operations Exercise 6.7

Question 1. Let’s calculate mentally 

1. 5x + 3x
Solution:

5x + 3x = 8x.

2. – 4y + 7y
Solution:

– 4y + 7y = 3y

3. 9y – 3y
Solution:

9y – 3y = 6y.

4. – 10x – 2x
Solution: –

10x-2x = – 12x

5. 3a + 4a – 2a
Solution:

3a + 4a – 2a = 3a – 2a

= 5a

6. – 7x-2x +5x
Solution:

– 7x-2x +5x = – 9x + 5x

= – 4x

7. 6p – 2p + 3p
Solution:

6p – 2p + 3p = 9p – 2p

= 7p

8. 4x² – 2x² – 3x² + x²
Solution:  

4x² – 2x² – 3x² + x²

= 5x² – 5x²

= 0

9. 5a²b – 2a²b – 3a²b + 8a²b
Solution:

5a²b – 2a²b – 3a²b + 8a²b = 13a²b – 5a²b

= 8a²b

10. 3x² – 6x² – 2x² – x² + 6x²
Solution: 

3x² – 6x² – 2x² – x² + 6x² = 9x² – 9x²

= 0

= 2.

Question 2. 

1. My age is ‘x’ years. Pallabi is 2 years older than me. Let’s find the sum of our ages.
Solution: 

Given

My age is ‘x’ years. Pallabi is 2 years older than me.

My age  = x years

Pallabi’s age = (x + 2) years

∴ Sum of our ages = (2x + 2) years

2. Today made ’x’ number of flower garlands. Mir made 6 more than twice the number of garlandsI made. In total, how many garlands we two have made.
Solution :

Given

Today made ’x’ number of flower garlands. Mir made 6 more than twice the number of garlandsI made.

I made = x  flower garlands

Mir made = 2x + 6 flower garlands

∴ Total number of flower

Garlands we made = 3x + 6

3. Today Ratul bought guava for Rs. x, apple for Rs. (x + 15) and cucumber for Rs. (2x + 3). Let’s find, how much money did Ratul spent today on fruits.
Solution:

Given

Today Ratul bought guava for Rs. x, apple for Rs. (x + 15) and cucumber for Rs. (2x + 3).

Ratul bought guava for = Rs. x

Apple for = Rs. (x + 15)

Cucumber for = Rs. (2x + 3)

∴ Ratul spent today on fruits = (x + x + 15 +2x + 3)

= Rs. (4x + 18)

4. Last year Firoza was present in school for x days. Firoza’s friend Mohini was present for (3x + 13) days, Let’s find, how much more days Mohini was present in school than Firoza last year.
Solution :

Given

Last year Firoza was present in school for x days. Firoza’s friend Mohini was present for (3x + 13) days

Mohini was present for = 3x + 13 days

Firoza was present for = x days

Mohini was present in school than Firoz = 3x + 13 = x

(3x + 13 –  x)  = 3x –  x+ 13

= (2x + 13) days

5. Today, Dipuda sold (2x + 19) newspapers. But yesterday he sold (5x – 8) newspapers. Let’s find, how many more newspapers did Dipuda sold today than yesterday.
Solution :

Given

Today, Dipuda sold (2x + 19) newspapers. But yesterday he sold (5x – 8) newspapers.

Yesterday he sold = (5x-8) newspaper

Today Dipuda sold = (2x+19) newspaper

∴ Dipuda sold more newspaper on yesterday  = (5x-8) = (2x+19)

(5x-8-2x-19) = 5x- 8- 2x -19

= 3x – 27

6. Pareshbabu earns Rs. 8x per month. But he spends Rs. (3x – 15) per month. Let’s find the amount of money he saves per month.
Solution :

Given

Pareshbabu earns Rs. 8x per month. But he spends Rs. (3x – 15) per month.

Monthly Incom of Pareshbabu = Rs. 8x

Monthly expenditure of Pareshbabu = Rs. 3x-15.

Monthly saving of Pareshbabu = 8x= 3x-15.

(8x – 3x +15 )=  8x – 3x+15 = 0

= 5x +15

Question 3. Let’s add:

1. 3a + b; 2a + 4b; 5a – b
Solution :

Given

3a + b; 2a + 4b; 5a – b

(3a +b) + (2a + 4b) + (5a – b)

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

= 1 0a + 4b

2. 5a – 4; 2a + 3; 2a – 4
Solution :

Given

5a – 4; 2a + 3; 2a – 4

(5a – 4) + (2a + 3) + (2a – 4)

= (5a + 2a+ 2a) + (- 4 + 3 – 4)

= 9a – 5

3. 6a + 7a +3; 9a² – 2a + 7; 4a² – 2a + 9
Solution :

Given

6a + 7a +3; 9a² – 2a + 7; 4a² – 2a + 9

(6a² + 7a + 3) + (9a² – 2a + 7) + (4a² – 2a + 9)

= (6a² + 9a² + 4a²) + (7a – 2a – 2a) + (3 + 7 + 9)

= 19a² + 3a + 19

4. 2a²b + 5b²a + 7; 3a²b – 2b²a + 6; 8a²b – b²a + 9
Solution:

Given

2a²b + 5b²a + 7; 3a²b – 2b²a + 6; 8a²b – b²a + 9

(2a²b + 5b²a + 7) + (3a²b – 2b²a + 6) + (8a²b – b²a + 9)

= (2a²b + 3a²b + 8a²b) + (5b²a – 2b²a – b²a) + (7 + 6 + 9)

= 13a²b + 2b²a + 22

5. 4xy + 5x + 7y; – 4xy – y – 3x; 3xy – 3y + 2x
Solution :

Given

4xy + 5x + 7y; – 4xy – y – 3x; 3xy – 3y + 2x

(4xy + 5x 7y) + (- 4xy y – 3x) + (3xy – 3y + 2x)

= (4xy – 4xy + 3xy) + (5x – 3x + 2x) + (7y – y – 3y)

= 3xy + 4x + 3y

Question 4. Let’s subtract:

1. (2x + 3y) from (8x + 6y)
Solution :

Given

(2x + 3y) and (8x + 6y)

(8x +6y) – (2x + 3y)

= 8x – 2x + 6y – 3y

= 6x + 3y

2. (m² – 2) from (- 3m² + 2m + 2)
Solution :

Given

(m² – 2) and (- 3m² + 2m + 2)

(- 3m² + 2m + 2) – (m² – 2)

= – 3m² – m² + 2m + 2 + 2

= – 4m² + 2m + 4

3. (8x + 4y + 7) from (2x + 3y)
Solution :

Given

(8x + 4y + 7) and (2x + 3y)

(2x + 3y) – (8x + 4y + 7)

= 2x – 8x + 3y – 4y – 7

= – 6x – y – 7

4.  (5a² + 2a – 1 ) from (- 9a² + 3a + 2)
Solution:

Given

(5a² + 2a – 1 ) and (- 9a² + 3a + 2)

(- 9a² + 3a + 2) – (5a² + 2a – 1 )

= -9a²-5a²+ 3a-2a + 2 + 1 .

= – 14a² + a + 3

5. (- 2x² + 3y² ) from x
Solution :

Given

(- 2x² + 3y² ) and x

x – (- 2x² + 3y²) = x + 2x² – 3y²

6. (2x² + xy + 3y²) from (3x² + 5xy)
Solution :

Given

(2x² + xy + 3y²) and (3x² + 5xy)

(3x² + 5xy) – (2x² + xy + 3y²)

= 3x² + 5xy – 2x² – xy – 3y²

= 3x² – 2x² + 5xy – xy – 3y²

= x² + 4xy – 3y²

Question 5. Let us Simplify the following :

1. 17x²y – 3xy² + 14x² y + 2xy²
Solution:

17x²y – 3xy² + 14x²y + 2xy²= 17x²y + 1 4 x²y – 3xy² + 2x²y

= 31x²y-xy²

2. – 5b + 18a + 6b – 2a
Solution :

– 5b + 1 8a + 6b – 2a = 1 8a – 2a + 6b – 5b

= 16a+b

3. 4m² + 3n² – (6m² + 7n²)
Solution :

4m² + 3n² – (6m² + 7n²)- 4m² – 6m² + 3n² – 7n²

= – 2m² – 4n²

4. a-b-(b-a)
Solution :

a – b – (b – a) = a – b – b + a

= a + a – b – b

= 2a – 2b

5.  (6p – 4q + 2r) + (2p + 3q – 4r)
Solution :

(6p – 4q + 2r) + (2p + 3q – 4r) = 6p + 2p – 4q + 3q + 2r – 4r

= 8p-q -2r

6.  – x + y + z – (2x – 3y + z)
Solution :

– x + y + z – (2x – 3y + z) = – x + y + z – 2x + 3y – z

= – x -2x + y + 3y + z – x

= – 3x + 4y (Ans )

7.  (x² + 2x – 5) + (3x² – 8x + 5)
Solution :

(x² + 2.x – 5) + (3×2 – 8x + 5) = x² + 2x – 5 + 3x² – 8x + 5

= x² + 3x² + 2x – 8x – 5 + 5

= 4x² – 6x

8. (7x² – 3x + 3) – (2x² – 13x – 7)
Solution :

(7x² – 3x + 3) – (2x² – 13x – 7) = 7x² – 3x + 3 – 2x² + 13x + 7

= 7x² – 2x² – 3x + 13x + 3 + 7

= 5x² + 1 0x + 1 0

9.  6a – 2a – ab – (3a + b – ab) + 2ab – b + a
Solution:

6a – 2a – ab – (3a + b – ab) + 2ab – b + a

= 6a – 2b – ab – 3a – b + ab + 2ab – b + a

= 6a – 3a + a – 2b – b – b – ab + ab + 2ab

= 4a – 4b + 2ab

Question 6. Ramu had Rs. (13x² + x – 3). He $pent Rs. (4x² – 3x – 12). Let’s find, how much money Ramu has got.
Solution :

Given

Ramu had Rs. (13x² + x – 3). He $pent Rs. (4x² – 3x – 12).

Ramu had =  Rs. 13x² + x – 3

He spent = Rs. 4x² – 3x -12

Remaining amount = Rs. (13x² + x – 3 – 4x² + 3x +12)

=   Rs. 9x² + 4x + 9

∴ Ramu has got Rs. 9x² + 4x + 9

Question 7. The lenght of three sides of a triangle are (x + 4) cm, (2x + 1) cm, and (4x – 8) cm, Let’s find the perimeter of the triangle.
Solution:

Given

The lenght of three sides of a triangle are (x + 4) cm, (2x + 1) cm, and (4x – 8) cm,

The perimeter of the triangle = Sum of the three side of the triangle.

= x + 4 + 2x + 1 +4x-8

= x + 2x + 4x + 4 + 1 – 8

= x + 2x + 4x + 5 – 8

= 7x – 3

Question 8. How much be added to – 8x² + 8x + 1 to get – 14x² + 11x – 3
Solution:

Required No. = (- 14x² + 1 1x – 3) – (- 8x² + 8x + 1)

= – 1 4x² + 11 x – 3 + 8x² – 8x – 1

= – 1 4x² + 8x² + 11 x – 8x – 3 – 1

= – 6x² + 3x – 4

Question 9. Let’s find, what must be subtracted from – 11 x – 7y – 9z to get – 7x +3y-5z.
Solution:

Required No. = (- 1 1x – 7y – 9z) -(- 7x + 3y – 5z)

= – 1 1 x – 7y – 9z + 7x – 3y + 5z

= – 1 1 x + 7x – 7y – 3y – 9z + 5z

= – 4x – 10y – 4z

Question 10. How much is the sum of (3x² + 4x) and (5x² – x) more than (3x – 5x²), let’s calculate.
Solution :

Required No. = (3x² + 4x) + (5x² – x) – (3x – 5x²)

= 3×2 + 4x + 5x² – x – 3x + 5x²

= 3x² + 5x² + 5x² + 4x – x – 3x

= 13x² + 4x – 4x

= 13x²

Question 11. Let’s subtract the sum (x² – 9x) and (- 2x² + 3x + 5) from the sum of (5 + 9x) and (6 – 7x + 4x²).
Solution : 

Required No. = {(5 + 9x) + (6 – 7x + 4x²)} – {(x² – 9x) + (- 2x² + 3x + 5)}

= (5 + 9x + 6 – 7x + 4x²) – (x² – 9x – 2x² + 3x + 5)

= (4x² + 9x – 7x + 5 + 6) – (x² – 9x – 2x² + 3x + 5)

= (4x² + 2x + 1 1 ) – (- x² – 6x + 5)

= 4x² + 2x + 1 1 + x² + 6x – 5

= 4x² + x² + 2x + 6x + 1 1 – 5

= 5x² + 8x + 6

WBBSE Class 7 Math Solution Algebraic Operations Exercise 6.8

Question 1. If x = 5 let’s find the values of the following algebraic expressions. When x = 5

1. 6x + 11
Solution:

6x+ 11 = 6 × 5 + 11 = 30 + 11 = 41.

2. \(\frac{x}{5}\) +2 
Solution:

⇒ \(\frac{x}{5}\)+2

= \(\frac{5}{5}\)+2

= 1+2 = 3

3. x² + 2x – 1
Solution :

x² + 2x – 1 = (5)²+ 2× 5-1

= 25 + 10-1

= 35- 1

= 34

4. x³> 8
Solution :

x³ + 8 = (5)³ + 8

= 125 + 8

= 133

5. 10 -x
Solution:

10 – x = 10 – 5

= 5

Question 2. If y= – 3, let’s find the values of the following algebraic expressions. \(\frac{y+5}{4}\)
Solution :

5-y = 5-(-3)

= 5 + 3

= 8

3. y + 8
Solution :

y + 8 = -3 + 8

= 5

4.  y² + 2y + 3
Solution :

y² + 2y + 3 = (- 3)² + 2 (- 3) + 3 =

9 – 6 + 3

= 12 – 6

= 6

5. y³ – 1
Solution :

= (-3)3 – 1

= – 27 – 1

= -28

Question 3. Let’s find the values of the following, when x = 2, y = – 1

1. 2x + 7y
Solution :

2x + 7y = 2 (2) + 7 (- 1)

= 4-7

= – 3

2x + 7y = – 3

2. x² + y²
Solution:

x² + y²= (2)² + (- 1)²

= 4 + 1

= 5

x² + y²  = 5

3. x² + 7xy + y²
Solution :

x² + 7xy + y²= (2)² + 7. 2 (- 1) + (- 1)²

= 4 – 14 +1

= – 14 + 5

= – 9

4. x³ – 8y³
Solution:

x³ – 8y³ = (2)³ – 8(- 1)³ = 8 – 8 (- 1)

= 8 + 8

= 16

5. \(\frac{x}{9}+\frac{y}{4}\)
Solution:

⇒ \(\frac{x}{9}+\frac{y}{4}\)

= \(\frac{2}{9}+\frac{-1}{4}\)

= \(\frac{8-9}{36}\)

= \(\frac{1}{36}\)

WBBSE Class 7 Math Solution Algebraic Operations Exercise 6.9

Question 1. Let’s find the product of the following.

1. 7,2x
Solution:

7,2x = 7 × 2x = 14x (Ans.)

2. – 3x, 4x
Solution:

– 3x, 4x = (- 3x) × (4x)

= – 12 x²

3. – 2x, – 3x²
Solution:

– 2x, – 3x²= (- 2x) × (- 3x²)

= 6x³

4. 7x, 0
Solution :

7x, 0 = 7x  × 0 = 0

5. 3ab, 4ac
Solution :

3ab, 4ac= (3ab) × (4ac)

= 12a²bc (Ans.)

6. 8x², 2y²
Solution:

8x², 2y² = (8x²) × (2y²)

= 16x²y²

7. 2a²b, 3ab²
Solution :

2a²b, 3ab² = 2a²b × 3ab²

= 6a³b³

8.  (- 4xy), (- 4xy)
Solution:

(- 4xy), (- 4xy)  = (- 4xy) × (- 4xy) = 16x²y²

Question 2. Let’s muliply first monomial with second monomial and write the product in corresponding blanks spaces.
Solution :

Algebraic Operations Exercise 6.10

Question 1. In each of the following cases, let’s multiply and find their product.

1.  ab, (a² – b²)
Solution:

ab, (a² – b²) = ab x (a² – b²)

= ab³ – ab³

2. 4a, (a + b – c)
Solution :

4a, (a + b – c)= 4a × (a + b – c)

= 4a² + 4ab – 4ac

3. 6a²b², (2a + b)
Solution :

6a²b², (2a + b) = 6a²b² × (2a + b)

= 12a³b² + 6a²b³ (Ans.)

4. xyz, (x²y – y²x + z²y)
Solution :

xyz, (x²y – y²x + z²y) = xyz x (x²y – y²x + z²y)

= x³y²z – x²y³z + xy²z³

5. 0, (ab + bc – ca)
Solution :

0, (ab + bc – ca) = 0 x (ab + be + ca)

= 0

Question 2. Let’s simplify

1.  7x (2x + 3) – 5x (3x – 4)
Solution :

7x (2x + 3) – 5x (3x – 4) = 7x (2x + 3) – 5x (3x – 4)

= (14x² + 21 x) – (1 5x² – 20x)

= 1 4x² + 21 x – 1 5x² + 20xy

= 14x² – 15x² + 21 x + 20x

= – x² + 41x

2. x(x – y) + y (y – z) + z (z – x)
Solution :

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

= x² + y² + z² – xy – yz – zx

3. 2x – 6x (5 – 8x – 3y)
Solution :

2x – 6x (5 – 8x – 3y) = 2x – 30x + 48x² + 18xy

= 48x² + 18xy – 28x

4. 7a – 2 (5a + 6b – 7)
Solution:

7a – 2 (5a + 6b – 7) = 7a – 10a – 12b + 14

= 14 -7a -12b.

WBBSE Class 7 Math Solution Algebraic Operations Exercise 6.11

Question 1. Let’s multiply :

1.  (10 – 3x) (7 + x)
Solution :

(10 – 3x) (7 + x) = 10 (7 + x) – 3x (7 + x)

= 70 + 10x – 21x – 3x²

= – 3x² – 1 1x + 70

2. (11 +2x) (8 – 2y)
Solution :

(11 + 2x) (8 – 2y) = (11+ 2x) × (8 – 2y)

= 11 (8 – 2y) + 2x (8 – 2y)

= 88 – 22y + 1 6x – 4xy

3. (a + by) (4a – 6y)
Solution :

(a + by) (4a – 6y) = (a + by) × (4a – 6y)

= a (4a – 6y) + by (4a – 6y)

= 4a² – 6ay + 4aby – 6by²

4. (2x² y – y² ) (3x-5y)
Solution :

(2x² y – y² ) (3x – 5y) = (2x² y – y² ) × (3 -5y)

= 2x² y (3x – 5y) – y² (3x – 5y)

= 6x³ y – 1 0x² y² – 3xy² + by³

5. \(\left(\frac{x}{2}-\frac{y}{3}\right)\left(\frac{2 x}{3}-\frac{3 y}{5}\right)\)
Solution:

⇒  \(\left(\frac{x}{2}-\frac{y}{3}\right)\left(\frac{2 x}{3}-\frac{3 y}{5}\right)=\left(\frac{x}{2}-\frac{y}{3}\right) \times\left(\frac{2 x}{3}-\frac{3 y}{5}\right)\)

= \(\frac{x}{2}\left(\frac{2 x}{3}-\frac{3 y}{5}\right)-\frac{y}{3}\left(\frac{2 x}{3}-\frac{3 y}{5}\right)\)

= \(\frac{x^2}{3}-\frac{3 x y}{10}-\frac{2 x y}{9}+\frac{y^2}{5}\)

= \(\frac{x^3}{3}-\frac{47 x y}{90}+\frac{y^2}{5}\)

6. \(\left(\frac{2 a^2}{7}-\frac{1}{5}\right)\left(\frac{3 a}{5}-\frac{2}{9}\right)\)
Solution:

⇒ \(\frac{2 a^2}{7}\left(\frac{3 a}{5}-\frac{2}{9}\right)-\frac{1}{5}\left(\frac{3 a}{5}-\frac{2}{9}\right)\)

=  \(\frac{6 a^3}{35}-\frac{4 a^2}{63}-\frac{3 a}{25}+\frac{2}{45}\)

Algebraic Operations Exercise 6.12

1. Let’s find values of the following mentally 

1. 3a × 4b 
Solution:

3a × 4b = 12ab

2. 12ab ÷ 3a 
Solution:

12ab ÷ 3a = 4b

3. 12ab ÷  3 
Solution:

12ab ÷  3 = 4ab or, 12ab ÷ 4ab = 3

4. (-x² ) × x 
Solution:

(-x² ) × x = – x³

5. 9x² – 3x² 
Solution:

9x² – 3x² = 3

6. x² × x² 
Solution:

x² × x² = x⁴

7. \(x^2 \times \frac{1}{x^2}\)
Solution:

\(x^2 \times \frac{1}{x^2}\)= 1

8.  0 ÷ ab = 0
Solution:

0 ÷ ab = 0

9. 4a²b²c² × 0
Solution:

4a²b²c² × 0 = 0

10. 3ab + 3b
Solution:

3ab + 3b = a or, = 3ab ÷ a = 3b.

11. x⁰ xy 
Solution:

x⁰ xy = 1 xy

= y

12. x÷ 0
Solution:

x÷ 0 =.undefined.

Question 2. Let’s multiply 

1. 2x² × (-3y) x 6z
Solution :

2x² × (-3y)× 6z = – 36x²yz

2. 7xy² × 8x²y × xy
Solution :

7xy² × 8x²y × xy= 56x⁴y⁴

3. (- 3a²) × (4a²b) × (-2)
Solution:

(- 3a²) × (4a²b) × (-2) = 24a⁴b

4. \((-2 m n) \times \frac{1}{6} m^2 n^2 \times 13 m^4 n^4\)
Solution: 

\((-2 m n) \times \frac{1}{6} m^2 n^2 \times 13 m^4 n^4\) = \((-2) \times \frac{1}{6} \times 13 m^{1+2+4} n^{1+2+4}\)

= \(\frac{13}{3} m^7 n^7\)

5. \(\frac{2}{3} x^2 y \times \frac{3}{5} x y^2\)
Solution: 

\(\frac{2}{3} x^2 y \times \frac{3}{5} x y^2\)

= \(\frac{2}{3} x \frac{3}{5} x^{2-1} y^{1+2}\)

= \(\frac{2}{5} x^3 y^3\)

6. \(\left(-\frac{18}{5} x^2 z\right) \times\left(-\frac{25}{6} x z^2 y\right)\)
Solution:

⇒ \(\left(-\frac{18}{5} x^2 z\right) \times\left(-\frac{25}{6} x z^2 y\right)\)

= \(\frac{18}{5} \times \frac{25}{6} x^{2+1} z^{1+2} y\)

= 15 x³z³y

7. \(\left(-\frac{3}{5} \mathrm{~s}^2 \mathrm{t}\right) \times\left(\frac{15}{7} \mathrm{st}^2 \mathrm{u}\right) \times\left(\frac{7}{9} \mathrm{su}^2\right)\)

⇒ \(\left(-\frac{3}{5} \mathrm{~s}^2 \mathrm{t}\right) \times\left(\frac{15}{7} \mathrm{st}^2 \mathrm{u}\right) \times\left(\frac{7}{9} \mathrm{su}^2\right)\)

= \(-\frac{3}{5} \times \frac{15}{7} \times \frac{7}{9} s^{2+1+1} t^{1+2} u^{1+2}\)

= – s⁴t³u³

8. \(\left(\frac{4}{3} x^2 y z\right) \times\left(\frac{1}{3} y^2 z x\right) \times\left(-6 x y z^2\right)\)
Solution:

⇒ \(\left(\frac{4}{3} x^2 y z\right) \times\left(\frac{1}{3} y^2 z x\right) \times\left(-6 x y z^2\right)\)

=  \(-\frac{4}{3} \times \frac{1}{3} \times 6 x^{2+1+1} y^{1+2+1} z^{1+1+2}\)

= – \(\frac{8}{3} x^4 y^4 z^4\)

9. 4a (3a + 7b)
Solution :

4a (3a + 7b) = 12a² + 28ab

10. 8a² × (2a + 5b)
Solution :

8a² × (2a + 5b)= 8a² × 2a+ 8a² x 5b

= 16 a³ + 40 a²b

11. – 1 7 x² × (3x – 4)
Solution :

– 1 7 x² × (3x – 4) = – 17x² × 3x – 17 x² (-4)

= – 51x³ + 68x²

12. \(\frac{2}{3} a b c\left(a^2+b^2-3 c^2\right)\)
Solution:

⇒ \(\frac{2}{3} a b c\left(a^2+b^2-3 c^2\right)\)

= \(\frac{2}{3} a b c \times a^2+\frac{2}{3} a b c \times b^2+\frac{2}{3} a b c\left(-3 c^2\right)\)

= \(\frac{2}{3} a^3 b c+\frac{2}{3} a b^3 c-2 a b c^3\)

13.  2 × 5x (10x²y – 100xy²)
Solution :

2 × 5x (10x²y – 100xy²) = 10x x (10x2y – 100xy2)

= 100x³y – 1000x²y²

14. (2x + 3y) (5x-y)
Solution :

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

= 10x²– 2xy + 15xy – 3y²

= 10 x² + 13xy – 3y²

15. (a² – b²) (2b – 6a)
Solution :

(a² – b²) (2b – 6a) = a² (2b²– 6a) – b² (2b – 6a)

= 2a²b – 6a³ – 2b³ + 6ab²

16. (x + 2.) (3x + 1 )
Solution :

(x + 2.) (3x + 1 ) = x (3x + 1 ) + 2 (3x + 1 )

= 3x² + x + 6x + 2

= 3x² + 7x + 2

Question 3.

1. Seema planted 3x saplings in a row, In 2x such rows let’s find how many saplings Seema can plant.
Solution :

In each row, Seema planted 3x saplings

∴ In 2x rows she will plant = 3x × 2x = 6x² saplings

2. The length of a rectangle is (4x + 1 ) m and its breadth is 3zm. Let’s calculate the area of the rectangle.
Solution :

Length & Breadth of the rectangle are (4x + 1 ) m & 3x m

∴ Area of the rectangle = (4x + 1) ×  3x sqm

= (12x²+3x) sqm.

3. Presently, the price of a dozen of bananas has increased by Rs. 6. If the previous price of a dozen of bananas was Rs. x, let’s find the cost of 2x dozen of bananas.
Solution :

⇒ Previously the price of one dozen banana was Rs. x.

⇒ Present price of one dozen banana is Rs. (x + 6)

⇒ Price of 2x dozen banana is Rs. (x + 6) × 2x

= Rs. (2x² + 12x).

4. Let’s find the area of a square whose each sode is 7x cm.
Solution :

Area of a square whose each side is 7x cm.

=  7x × 7x sqcm

= 49x² sqcm

5. The area of a rectangle is 8x² sq units. If its length is 4x units, let’s find its breadth.
Solution:

The area of rectangle = 8x² sq unit

And  its length = 4x unit

The breadth of the rectangle = (8x² ÷ 4x) unit

= 2x unit

6. Sushobhan sold 729y4 number of kites in 9y days. Let’s find the number of kites sold in average per day.
Solution:

Susobhan sold in 9y day 729 y⁴ kites

∴ He sold in 1 day = \(\frac{729 y^4}{9 y}\)  kites

= 81 y³ kites

4. Let’s divide the first algebraic expression by the second algebraic expression in the following pairs of algebraic numbers 

1. 8x³b, x²b
Solution :

= 8x³b ÷  x²b

= \(\frac{8 x^3 b}{x^2 b}\)

= 8x

2. 9xy³, xy
Solution:

9xy³, xy = – 9xy³ xy

= \(\frac{-9 x y^3}{x y}\)

= -9y²

3. -15xYz²,-x²yz²
Solution :

-15xyz²,-x²yz²= (- 1 5x²y⁴z²) -4- (- x²yz²)

= \(\frac{-15 x^2 y^4 z^2}{-x^2 y z^2}=15 y^3\)

4. 21 l³m³n³, – 4l⁴mn
Solution:

21 l³m³n³, – 4l⁴mn = (21 l³m³n³) ÷ (4l⁴mn)

= \(\frac{\left.21\right|^3 m^3 n^3}{-\left.4\right|^4 m n}\)

= \(-\frac{21}{4l} m^2 n^2\)

2. (5a² – 7ab²), a
Solution : = (5a² – 7ab²) ÷ a

= \(\frac{5 a^2}{a}-\frac{7 a b^2}{a}\)

= \(5 a-7 b^2\)

6. (- 48x⁹ + 12x⁶), 3x³
Solution :

(- 48x⁹ + 12x⁶), 3x³ = (- 48x⁹ + 1 2x⁶) ÷3x³

= \(\frac{-48 x^9}{3 x^3}+\frac{12 x^6}{3 x^3}\)

= – 16 x⁶ + 4x³

7. 15m²n + 20m²n², 5mn
Solution:

15m²n + 20m²n², 5mn = (15m²n + 20m²n²) ÷ (5mn)

= \(\frac{15 m^2 n}{5 m n}+\frac{20 m^2 n^2}{5 m n}\)

= 3m + 4mn

8. 36a⁵b² – 24a²b5, – 4a²b²
Solution:

36a⁵b² – 24a²b5, – 4a²b² = (36a⁵b² – 24a²b⁵) ÷ (-4a²b²)

= \(\frac{36 a^5 b^2}{-4 a^2 b^2}-\frac{24 a^2 b^5}{-4 a^2 b^2}\)

9. 3pqr + 6p²qr² – 9p³q²r³, – 3pqr
Solution :

3pqr + 6p²qr² – 9p³q²r³, – 3pqr = (3pqr + 6p²qr² – 9p³q²r³) ÷(- 3pqr)

= \(\frac{3 p q r}{-3 p q r}+\frac{6 p^2 q r^2}{-3 p q r}-\frac{9 p^3 q^2 r^3}{-3 p q r}\)

= – 1 – 2pr + 3²2qr²

10. m²n⁴ + m³n³ – m⁴n², – m⁴n⁴
Solution :

m²n⁴ + m³n³ – m⁴n², – m⁴n⁴ = (m²n⁴ + m³n³ – m⁴n²) ÷ (m⁴n⁴)

= \(\frac{m^2 n^4}{-m^4 n^4}+\frac{m^3 n^3}{-m^4 n^4}-\frac{m^4 n^4}{-m^4 n^4}\)

= \(-\frac{1}{m^2}-\frac{1}{m n}+\frac{1}{n^2}\)

5. Let us simplify the following –

1. a (b – c) + b (c – a) + c (a – b)
Solution :

a (b – c) + b (c – a) + c (a – b) a (b – c) + b (c – a) + c (a – b) = ab -ac +bc – ab + ac – bc

= 0

2. a (b – c) – b (c – a) – c (a – b)
Solution :

a (b – c) – b (c – a) – c (a – b) = ab – ac – bc + ab – ac + bc

= 2ab – 2ac .

3.  x (x + 4) + 2x (x – 3) – 3x²
Solution :

x (x + 4) + 2x (x – 3) – 3x² = x² + 4x + 2x² – 6x² – 3x

= x² – 2x – 3x² + 4x – 6x = – 2x

4. 3x² + x (x + 2) – 3x (2x + 1 )
Solution :

3x² + x (x + 2) – 3x (2x + 1 ) = 3x² + x² + 2x – 6x² – 3x

= 4x² – 6x² + 2x – 3x = – 2x² – x

5. (a + b) (a – b) + (b + c) (b – c) + (c + a) (c – a)
.Solution :

(a + b) (a – b) + (b + c) (b – c) + (c + a) (c a)

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

= a² – ab + ab – b² + b² – bc + bc – c² + c² – ac + ac – a²

= 0

6. (a² + b²) (a² – b²) + (b² + c²) (b² – c²) + (c² + a²) (c² – a²)
Solution :

(a² + b²) (a² – b²) + (b² + c²) (b² – c²) + (c² + a²) (c² – a²)

= a² (a² – b²) + b² (a² – b²) + b² (b² – c²) + c² (b² – c²) + c² (c² – a²) + a² (c² – a²)

= a⁴ – a²b² + a²b² – b⁴ + b⁴ – b²c² + b²c² – c⁴ + c⁴ – a²c² + a²c² – a⁴

= 0