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If x(y+z)=32,y(z+x)=65,z(x+y)=77 then find the value of xyz

If x(y+z)=32,y(z+x)=65,z(x+y)=77 then find the value of xyz

Grade:12th pass

1 Answers

Subhash
14 Points
6 years ago
x(y+z)=32
xy+xz=32 …...(1)
y(z+x)=65
yz+xy=65 …...(2)
z(x+y)=77
xz+xy=77 …..(3)
adding eq. 1,2&3--
2(xy+xz+yz)=174
xy+xz+yz=87 …..(4)
subs. eq. 1,2&3 from eq. 4 respectively, we get---
yz=55 …....(5)
xz=22  …...(6)
xy=10 …....(7)
therefore-   z=55/y  from eq. 5
and            x=10/y   from eq. 7
since xz=22   (eq. 6)
= 55/y * 10/y = 22
= 22 y2 = 55*10
= y2 = 550/22=25
y=5
x= 10/y = 10/5= 2
z= 55/y = 55/5 = 11
therefore--  xyz = 2*11*5 =110
therefore--  xyz=110

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