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Question

# Solve the following equations:x2+xy+xz=18,y2+yz+yz+12=0z2+zx+zy=30.

A
x=±2,y=2,z=±4
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B
x=±3,y=2,z=±5
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C
x=±4,y=±5,z=±5
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D
None of the above
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Solution

## The correct option is B x=±3,y=∓2,z=±5x2+xy+yz=18⇒x(x+y+z)=18 ........(i)y2+yz+yx+12=0⇒y(x+y+z)=−12 .......(ii)z2+zx+zy=30⇒z(x+y+z)=30 ........(iii) Dividing (i) by (ii) ⇒xy=−1812=−32y=−23x .......(a) Dividing (i) by(iii) xz=1830=35z=53x ...........(b) Substituting (a) and (b) in (i) x(x−23x+53x)=18⇒6x23=18⇒x2=9⇒x=±3 Substituting x in (a), we get ⇒y=−23(±3)=∓2 Substituting x in (b), we get ⇒z=53(±3)=±5 So, x=±3,y=∓2,z=±5

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