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
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


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





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

@ 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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