Question

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

Answer 1

Answer:

$2 {y}^{2} + 3yz + ( - {y}^{2} - y - {z}^{2} ) + yz + 2 {z}^{2} \: \: - (3 {y}^{2} - {z}^{2} + ( - {y}^{2} + yz + {z}^{2} ))$

$2 {y}^{2} + 3yz - {y}^{2} - y - {z}^{2} + yz + 2 {z}^{2} \: \: - (3 {y}^{2} - {z}^{2} - {y}^{2} - yz - {z}^{2} )$

$2 {y}^{2} + 3yz - {y}^{2} - y - {z}^{2} + yz + 2 {z}^{2} \: \: - 3 {y}^{2} - {z}^{2} - {y}^{2} - yz - {z}^{2}$

[/tex]

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

$2 {y}^{2} - 5 {y}^{2} + 4yz - yz + 2 {z}^{2} - 3 {z}^{2}$

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

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