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sunny: i have two equations and 3 variables x+2*y+3*z=0 and 2*x+ 3*y+3*z=0... i.e. infinite solutions .right? i tried reporting my answer in parametric form but by two different methods i got two different parametric solution i.e.(3*t,-3*t,t) and (-p,2p,-p). for any value of p or t the equations are satisfied.. why is it happening.
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  • January 4, 2017 at 4:51pm
  • Little Sky:Your 1st set of solution is valid but your 2nd set is not. Here is why: 1(3t) + 2(-3t) + 3(t) = 0 2(3t) + 3(-3t) + 3(t) = 0 [ieqn]\therefore[/ieqn]1st set of solution is valid. 1(-p) + 2(2p) + 3(-p) = 0 2(-p) + 3(2p) + 3(-p) = p (invalid) [ieqn]\therefore[/ieqn]2nd set of solution is not valid.
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    • January 6, 2017 at 4:22am
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