Question

Solve this please please, the question is in the attachment.... both parts pleaseee!

Answer 1

Answer:

1. 2x²-3y²+2x²-5y²+6x²+7y²

2x²+2x²+6x²-3y²-5y²+7y²

10x²-y² ans

2.a²+b²-2ab-(a²+b²+2ab)

a²+b²-2ab-a²-b²-2ab

a²-a²+b²-b²-2ab-2ab

-4ab ans

Answer 2

$\huge\bf\boxed{Hola \:Mate!}}$

A)

$\huge{The\: sum\: of \: \:2x^{2} - 3y^{2} , 2x^{2}-5y^{2}\: and \: 6x^{2} + 7y^{2}$

$= 2x^{2} - 3y^{2} + 2x^{2}-5y^{2} + 6x^{2} + 7y^{2}$

✩ Upon grouping the like terms together, we get,

$= 2x^{2} + 2x^{2} + 6x^{2}- 3y^{2} -5y^{2} + 7y^{2}$

✩ Then we add the like terms,

$= 10x^{2} - 1y^{2}$

$= 10x^{2} - y^{2}$

✩ So, therefore,

$\huge\red{The\: sum\: of \: \:2x^{2} - 3y^{2} , 2x^{2}-5y^{2}\: and \: 6x^{2} + 7y^{2}\:\: is$ $equal\: to\: 10x^{2} - y^{2}$

B)

$\huge{Subtraction \: of \: (a^{2} + b^{2} + 2ab) \: from \: (a^{2} + b^{2} - 2ab)$

$=(a^{2} + b^{2} - 2ab) -( a^{2} + b^{2} + 2ab)$

✩ First, we open the brackets,

$=a^{2} + b^{2} - 2ab - a^{2} - b^{2} - 2ab$

✩ Then, we group the like terms together,

$=a^{2} - a^{2} + b^{2} - b^{2} - 2ab - 2ab$

✩ And get,

$=0 + 0 -4ab$

$= -4ab$

✩ So, therefore,

$\huge\red{Subtraction \: of \: (a^{2} + b^{2} + 2ab) \: from \: (a^{2} + b^{2} - 2ab)\: gives \: -4ab}}$

♦ Remember

While opening the brackets of the second term in B), the operations were changed. This is an essential rule. Examples :-

✶2ab + 2z - ( 6ab - 7z )

  = 2ab + 2z - 6ab + 7z

✶9c + 2b + 3a - ( 4a - 2c + 3c )

 = 9c + 2b + 3a - 4a + 2c - 3c

HOPE I HELPED ..!! :)