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

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3 years ago

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