Question

25.
If x2 – 2y =-13, y2 – 4z = 14, z2 +6X =-15, then find the value of xy + x2 + 6yz​

Answer 1

xy + x² + 6yz = 18

Step-by-step explanation:

Given,

x² - 2y = - 13

y² - 4z = 14

z² + 6x = - 15

Adding we get

x² - 2y + y² - 4z + z² + 6x = - 13 + 14 - 15

or, x² + 6x + y² - 2y + z² - 4z = - 14

or, (x² + 6x + 9) + (y² - 2y + 1) + (z² - 4z + 4) = - 14 + 9 + 1 + 4

or, (x + 3)² + (y - 1)² + (z - 2)² = 0

Since the sum of the three positive terms is zero, we can conclude that each of them equals to zero.

Then, x + 3 = 0 i.e., x = - 3

y - 1 = 0 i.e., y = 1

z - 2 = 0 i.e., z = 2

Now, xy + x² + 6yz

= (- 3) * 1 + (- 3)² + 6 * 1 * 2

= - 3 + 9 + 12

= 18