## Precalculus (6th Edition) Blitzer

We can eliminate x from equations $1$ and $2$ by multiplying equation $1$ by $-2\text{ }$ and adding the equations. We can eliminate x from equations 1 and 3 by multiplying equation $1$ by $-4$ and adding the equations.
Let us consider the equations: $x+y-z=-1$ Equation 1 $2x-2y-5z=7$ Equation 2 $4x+y-2z=7$ Equation 3 Now, eliminate x from equations (I) and (II) by multiplying (I) by $-2$ to get; \begin{align} & -2x+\left( -2 \right)y+2z=-1\left( -2 \right) \\ & -2x-2y+2z=2 \\ \end{align} Then, add this equation with equation (II) to get; \begin{align} & -2x-2y+2z+\left( 2x-2y-5z \right)=2+7 \\ & -4y-3z=9 \end{align} Therefore, x is eliminated from equations (I) and (II). Then, eliminate x from equations (I) and (III) by multiplying (I) by $-4$ to get; \begin{align} & x\cdot \left( -4 \right)+y\cdot \left( -4 \right)-z\cdot \left( -4 \right)=-1\cdot \left( -4 \right) \\ & -4x-4y+4z=4 \end{align} Then, add this equation with (III) to get; \begin{align} & -4x-4y+4z+4x+y-2z=4+7 \\ & -3y+2z=11 \end{align} Thus, it is clear from this equation that x is eliminated from equations (I) and (III).