Question

If x, y, z are three integers such that x + y = 12, y + z = 16, z + x = 20. Find xyz.
a) 384
b) 392
c) 482
d) 472

Answer 1

b). 392.

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Answer 2

Answer:

$384$

Step-by-step explanation:

x + y = 12 -------> i

y + z = 16 -------> ii

z + x = 20 --------> iii

add i, ii, iii

2(x + y + z) = 48

x + y + z = 24 -------> iv

from i and iv

z = 12

from ii and iv

x = 8

from iii and iv

y = 4

therefore,

xyz = 8×4×12 = 384

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