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{[x-2y+z=2],[2x+2y-3z=-4],[5x+z=1]:}

Find xx,yy,zz values from the equations {x−2y+z=22x+2y−3z=−45x+z=1 \left\{\begin{array}{l}x-2 y+z=2 \\ 2 x+2 y-3 z=-4 \\ 5 x+z=1\end{array}\right.

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Q. Find xx,yy,zz values from the equations {x−2y+z=22x+2y−3z=−45x+z=1 \left\{\begin{array}{l}x-2 y+z=2 \\ 2 x+2 y-3 z=-4 \\ 5 x+z=1\end{array}\right.
  1. Identify Equations: Identify the system of equations:\newlinex−2y+z=2,2x+2y−3z=−4,5x+z=1. \begin{align*} x - 2y + z &= 2, \\ 2x + 2y - 3z &= -4, \\ 5x + z &= 1. \end{align*}
  2. Eliminate Variable: Use elimination to simplify the equations. Start by eliminating yy from the first two equations:\newlineMultiply the first equation by 22:\newline2x−4y+2z=4. 2x - 4y + 2z = 4. \newlineNow add this to the second equation:\newline(2x−4y+2z)+(2x+2y−3z)=4−4. (2x - 4y + 2z) + (2x + 2y - 3z) = 4 - 4. \newline4x−2z=0. 4x - 2z = 0.
  3. Simplify Equation: Simplify the new equation:\newline4x−2z=0⇒2x=z. 4x - 2z = 0 \Rightarrow 2x = z. \newlineSubstitute 2x2x for zz in the third equation:\newline5x+2x=1⇒7x=1⇒x=17. 5x + 2x = 1 \Rightarrow 7x = 1 \Rightarrow x = \frac{1}{7}.
  4. Find x: Substitute x=17x = \frac{1}{7} back into 2x=z2x = z:\newlinez=2×17=27. z = 2 \times \frac{1}{7} = \frac{2}{7}.
  5. Find z: Substitute x=17x = \frac{1}{7} and z=27z = \frac{2}{7} into the first equation to find yy:\newline17−2y+27=2. \frac{1}{7} - 2y + \frac{2}{7} = 2. \newline−2y+37=2⇒−2y=2−37⇒−2y=117. -2y + \frac{3}{7} = 2 \Rightarrow -2y = 2 - \frac{3}{7} \Rightarrow -2y = \frac{11}{7}. \newliney=−1114. y = -\frac{11}{14}.

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