Question

please answer the above​

Answer 1

$\LARGE{ \underline{ \purple{ \sf{Required \: answer:}}}}$

Here we are provided with two equations, and we have to find the value of z accordingly.

Equation (1):

$\sf{ \dfrac{ {x}^{2} + {y}^{2} + {z}^{2} - 64 }{xy - yz - z} = - 2}$

Equation (2):

$\sf{x + y = 3z}$

We have to find the value of z....?

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Starting with eq. (1),

Cross multiplying:

➝ x² + y² + z² - 64 = -2xy + 2yz + 2zx

Now bringing the RHS elements to the LHS and shifting -64 to the RHS,

➝ x² + y² + z² + 2xy - 2yz - 2zx = 64

This can be written as,

➝ x² + y² + (-z)² + 2xy + 2y(-z) + 2(-z)x = 64

By using the identity for finding square of a trinomial:

➝ (x + y - z)² = 64

That means,

➝ x + y - z = $\pm$8 ----------(3)

Now from Equation (2) and Equation (3),

1️⃣ If it is equals to +8

• x + y - z = +8
• x + y - 3z = 0

Then, z = 4 (By subtracting the equations)

2️⃣ If it is equals to -8

• x + y - z = -8
• x + y - 3z = 0

Then, z = -4 (Again by subtracting)

According to the options given:

$\therefore$ The correct answer is 4 (Option C)