Math, asked by deepakdeeps2007, 2 months ago

solve the simultaneous equation 3x-2y+z=0 ,4x+6y-3z=13,x-2y+2z=-4​

Answers

Answered by MaheswariS
3

\textbf{Given:}

\mathsf{Equations\;are}

\mathsf{3x-2y+z=0}

\mathsf{4x+6y-3z=13}

\mathsf{x-2y+2z=-4}

\textbf{To find:}

\textsf{Solution of the given equations}

\textbf{Solution:}

We apply elimination method to solve the given equations

\textsf{Consider,}

\mathsf{3x-2y+z=0}------(1)

\mathsf{4x+6y-3z=13}--------(2)

\mathsf{x-2y+2z=-4}------------(3)

\mathsf{(1)-3{\times}(3):}

\mathsf{3x-2y+z=0}

\mathsf{3x-6y+6z=-12}

\mathsf{4y-5z=12}------------(4)

\mathsf{(2)-4{\times}(3):}

\mathsf{4x+6y-3z=13}

\mathsf{4x-8y+8z=-16}

\mathsf{14y-11z=29}------------(5)

\mathsf{7{\times}(4)-2{\times}(5):}

\mathsf{28y-35z=84}

\mathsf{28y-22z=58}

\mathsf{-13z=26}

\mathsf{z=\dfrac{26}{-13}}

\implies\boxed{\mathsf{z=-2}}

\mathsf{Put\;z=-2\;in\;(4),\;we\;get}

\mathsf{4y-5(-2)=12}

\mathsf{4y+10=12}

\mathsf{4y=12-10}

\mathsf{4y=2}

\mathsf{y=\dfrac{2}{4}}

\implies\boxed{\mathsf{y=\dfrac{1}{2}}}

\mathsf{(3)\implies}

\mathsf{x-2y+2z=-4}

\mathsf{x-2{\times}\dfrac{1}{2}+2(-2)=-4}

\mathsf{x-1-4=-4}

\mathsf{x=-4+5}

\implies\boxed{\mathsf{x=1}}

\therefore\underline{\mathsf{Solution\;is}}

\mathsf{x=1}

\mathsf{y=\dfrac{1}{2}}

\mathsf{z=-2}

\textbf{Find more:}

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