subtract the sum of

z square + yz - y square and 3y square - z square from the sum of yz + 2z square , 2y square + 3y square, -y square - z square - yz

what is the answer of this expression ?

 

Now, Subtract (i) from (ii), we get,

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[yz+2y^2+3y^2+(-y^2-z^2-yz)] - [z^2 +yz-y+(3y^2-z)]

 

=[yz+2y^2+3y^2+y^2-z^2-yz] - [z^2 +yz-y+3y^2-z]

=[6y^2-z^2] - [z^2 +yz-y+3y^2-z}
 
=6y^2-z^2 - z^2-yz+y-3y^2+z
 
=3y^2-2z^2-yz+y+z
 
Hope it helps
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{ [ (yz+2z2)+(2y2+3y2)+(y2-z2-yz) ] – [ ( z2+yz-y2)+3y2 ] }

 

Let us solve the first square bracket first...

 

yz+2z2 +2y2 +2y2+3y2-y2+z2+yz

= yz+yz+2z2+z2+2y2+2y2+3y2+y2

=2yz+3z2+8y2

 Now, let us solve the whole equation....

=(2yz+3z2+8y2)-(Z2+yz-y2+3y2)

=2yz+3z2+8y2-z2-yz+y2-3y2

=2yz+yz+3z2-z2+8y2+y2-3y2

=3yz+2z2+6y2

Therefore, =3yz+2z2+6yis the answer.

@ eshan.........in the forst step, i.e.[yz+2y^2+3y^2+y^2-z^2-yz], you did not change the sighnature for the bold one... 'cause negative and positive make up negative....

Hope this helps you.........

If this helps, please give a thumbs up........ :-D

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