Question

from the sum of 2y²+2yz, -y²-yz-z² and yz +2z², subtract the sum of 3y²-z² and -y²+yz+z²

Answer 1

Answer:

If you don't have a expressions just put it 0

I didn't put it because you will confuse again soo....

Answer 2

Answer:

the answer is  $-3y^{2} -z^{2}$

Step-by-step explanation:

sum of       $2y^{2} + 2yz , -y^{2} -yz - z^{2}$

                  = $2y^{2} + 2yz -y^{2} -yz -z^{2}$

                  =$y^{2} +yz-z^{2}$  

 sum of   $3y^{2} - z^{2} , y^{2} +yz +z^{2}$

            = $3y^{2} -z^{2} +y^{2} +yz +z^{2}$

              $= 4y^{2} +yz$  

subtract  them

            = $y^{2} + yz - z^{2} -[4y^{2} + yz]$

             = $y^{2} + yz -z^{2} -4y^{2} } -yz$

             =$-3y^{2} - z^{2}$