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Question

Solve the following equations:
3x+y−2z=0,
4x−y−3z=0,
x3+y3+z3=467

A
x=5;y=1;z=5
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B
x=2;y=1;z=7
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C
x=3;y=2;z=7
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D
x=5;y=1;z=7
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Solution

The correct option is D x=5;y=1;z=7
Let 3x+y2z=0 ........(i)
4xy3z=0 .........(ii)
and x3+y3+z3=467 .......(iii)
Solving (i) and (ii) by cross multiplication
x32=y8+9=z34=k (let)
x5=y1=z7=kx=5k,y=k,z=7k .........(iv)
Substituting x,y,z in (iii), we have
(5k)3+(k)3+(7k)3=467125k3+k3+343k3=467467k3=467k=1
Substituting k in (iv)
x=5,y=1,z=7
which is the required solution

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