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Elimination Method 2x-y-5z=27-2x-4y+7z=-27-2x-6y+3z= -3

User Ram Mourya
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The given system of linear equations is


\mleft\{\begin{aligned}2x-y-5z=27 \\ -2x-4y+7z=-27 \\ -2x-6y+3z=-3\end{aligned}\mright.

Let's sum the first and the second equation to eliminate x, and to get an equation with y and z only.


\begin{gathered} 2x-2x-y-4y-5z+7z=27-27 \\ -5y+2z=0 \end{gathered}

Then, we sum the first and the third equation to eliminate x, and to get an equation with y and z only.


\begin{gathered} 2x-2x-y-6y-5z+3z=27-3 \\ -7y-2z=24 \end{gathered}

Now, we form a system with the new two equations.


\mleft\{\begin{aligned}-5y+2z=0 \\ -7y-2z=24\end{aligned}\mright.

Let's sum the equations to eliminate z and find y.


\begin{gathered} -5y-7y+2z-2z=0+24 \\ -12y=24 \\ y=(24)/(-12) \\ y=-2 \end{gathered}

y is equal to -2.

Then, we use the y-value to find z.


\begin{gathered} -5y+2z=0 \\ -5(-2)+2z=0 \\ 10+2z=0 \\ 2z=-10 \\ z=-(10)/(2) \\ z=-5 \end{gathered}

z is equal to -5.

Now, we use y and z values to find x.


\begin{gathered} 2x-y-5z=27 \\ 2x-(-2)-5(-5)=27 \\ 2x+2+25=27 \\ 2x=27-25-2 \\ 2x=0 \\ x=(0)/(2) \\ x=0 \end{gathered}

x is equal to zero.

Therefore, the solution to the system is (0, -2, -5).

User Colm Bhandal
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