Question

Q 10) find each of the following product
1. (2x-3y) (2x-3y) 2. (3x-2y) (2x-3y)
3. (1/2x-1/5y) (1/2x-1/5y)
4. (3x+2y) (3x-2y)​

Answer 1

Step-by-step explanation:

Given:-

1. (2x-3y) (2x-3y)

2. (3x-2y) (2x-3y)

3. (1/2x-1/5y) (1/2x-1/5y)

4. (3x+2y) (3x-2y)

To find:-

find each of the following product ?

Solutions:-

1.Given expression is (2x-3y) (2x-3y)

=> (2x-3y)^2

This is in the form of (a-b)^2

Where , a = 2x and b = 3y

We know that

(a-b)^2 = a^2 - 2ab + b^2

(2x-3y)^2=>

(2x)^2 - 2(2x)(3y) + (3y)^2

=> 4x^2 - 12xy +9y^2

(2x-3y) (2x-3y) = 4x^2 - 12xy +9y^2

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2.Given expression is (3x-2y) (2x-3y)

=> 3x(2x-3y) -2y(2x-3y)

=> (3x)(2x)-(3x)(3y) -(2y)(2x) -(2y)(-3y)

=> 6x^2 -9xy -4xy +6y^2

=> 6x^2 -13xy +6y^2

(3x-2y) (2x-3y) = 6x^2 -13xy +6y^2

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3.Given expression is (1/2x-1/5y) (1/2x-1/5y)

It can be written as (1/2 x - 1/5 y)^2

This is in the form of (a-b)^2

Where , a = 1/2 x and b = 1/5 y

We know that

(a-b)^2 = a^2 - 2ab + b^2

(1/2 x - 1/5 y)^2=>

=> (1/2 x)^2 -2(1/2 x) (1/5 y ) +(1/5 y )^2

=> (1/4 x ^2 )- 2(1/2)(1/5)(xy) + (1/25 y^2)

=>(1/4 x^2 - (1/5 xy ) + (1/25 y^2)

or

=> x^2/4 - xy /5 + y^2/25

or

LCM of 4,25,5 is 100

=> (25x^2-20xy+4y^2)/100

(1/2x-1/5y) (1/2x-1/5y)

=1/4 x^2-1/5 xy +1/25 y^2

(or) x^2/4 - xy /5 + y^2/25

(or) (25x^2-20xy+4y^2)/100

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4.Given expression is (3x+2y) (3x-2y)

It is in the form of (a+b)(a-b)

Where , a = 3x and b = 2y

we know that

(a+b)(a-b)=a^2-b^2

Now

(3x+2y) (3x-2y)=>

=> (3x)^2 - (2y)^2

=> (3x×3x) - (2y×2y)

=> 9x^2-4y^2

(3x+2y)(3x-2y)= 9x^2 - 4y^2

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Used formulae:-

• (a-b)^2 = a^2 -2ab +b^2

• (a+b)(a-b)=a^2-b^2