Question

Xyz+ xy + xz + yz + x + y + z =384then x+y+Z=?

Answer 1

xyz +xy + yz + xz +x + y + z = 384

=> xy( z+1) + z( x + y) + (x+y) + z = 384

=> xy(z+1) +( x+y)(z+1) + z = 384

=> xy(z+1) + (x+y)(z+1) +( z+1) = 384 +1 (added 1 to both the sides)

=> (z+1) ( xy + x+y + 1) = 385

=> ( z+1) {x(y+1) + 1( y+1)} = 385

=> ( z+1)(y+1)( x+1) = 5 x 7 x 11

=> (x+1), (y+1), (z+1) hold either of these values. 5 or 7 or 11

=> x, y, z hold either of the values 4, 6 , or 10

Hence, x+y+ z = 4+6+10= 20ans..

Answer 2

hey

gd mrng

here is your answer

xyz+xy+xz+yz+x+y+z=384

xy (z+1) + z (x+y)+1(x+y)+z=384

xy (z+1) + (x+y) (z+1) +(z+1)−1=384

xy (z+1) + (x+y) (z+1) +(z+1)=385

(z+1) (xy+x+y+1)=385

(z+1) (x(y+1)+1(y+1))=385

(x+1)(y+1)(z+1)=385

(x+1)(y+1)(z+1)=5×7×11

On comparing ,x+1 = 5 , y+1=7 , z+1=11or x+1 = 7 , y+1=11 , z+1=5or x+1 = 11 , y+1=5 , z+1=7 Now ,  

For each case

x+1+y+1+z+1=5+11+7

x+y+z+3=23

x+y+z=20

hope its help you

plz mark as brainlist answer

plz follow me for more brainlist answers

thanxx and be brainly..............