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Question

# x−2y+z=0y−z=22x−3z=10 Find the value of x,y,z

A
x=1;y=0;z=2
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B
x=1;y=0;z=1
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C
x=1;y=1;z=2
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D
x=2;y=0;z=2
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Solution

## The correct option is C x=2;y=0;z=−2Given, x−2y+z=0→(1)y−z=2→(2)2x−3z=10→(3)From equation 2y=2+zSubstitute this 'y' value in equation (1)x−2y+z=0→1x−2(2+z)+z=0x−4−2z+z=0x−4−z=0x−z=4→(4)Now from equation (3) and equation (4)2x−3z=10→(3)x−z=4→(4)Multiply (3) by 1 and (4) by 2 and subtract both equations2x−3z=10→32x−2z=8→4−+−−z=2z=−2Substitute 'z' value in equation (3)2x−3(−2)=10x=2Substitute 'z' value in equation (2)We get y−(−2)=2y=0∴By solving the given three equations we getx=2;y=0;z=−2

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