Question

14. From the sum of 2y2 + 3yz, - y - yz – 22 and yz + 2z2, subtract the sum of 3y - 22
and - y2 + y2 + z2​

Answer 1

Answer:

sorry I didn't know that answer

sorry..!

Answer 2

$sum \: of \: 2 {y}^{2} + 3yz \: \: and \: - y - yz - 22 \: and \: yz + 2 {z}^{2} \\ = 2 {y}^{2} + 3yz - y - yz - 22 + yz + 2 {z}^{2} \\ = 2 {y}^{2} + 2 {z}^{2} - y - 22 + 3yz - - - (1) \\ sum \: of \: 3y - 22 \: and \: - {y}^{2} + {y}^{2} + {z}^{2} \\ = 3y - 22 - {y}^{2} + {y}^{2} + {z}^{2} \\ = 3y + {z}^{2} - 22 - - - (2) \: \\ from \: \: 2{y}^{2} + 2 {z}^{2} - y - 22 + 3yz \: subtract \: 3y + {z}^{2} - 22 \\ = 2 {y}^{2} + 2 {z}^{2} - y - 22 + 3yz - 3y - {z}^{2} + 22 \\ = 2 {y}^{2} + {z}^{2} - 4y + 3yz$

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