## Precalculus (6th Edition) Blitzer

The solution of the system of equations is $x=7,y=4\text{ and }z=5$.
Let us consider the given equation: $x+y+z=16$ …… (I) $2x+3y+4z=46$ …… (II) $5x-y=31$ …… (III)\ Equation (I): multiply by $4$ and substrate from equation (II): \begin{align} & 2x+3y+4z=46 \\ & 4x+4y+4z=64 \\ & \underline{-\text{ }-\text{ }-\text{ }-} \\ & -2x-y=-18 \\ \end{align} Further simpliy: $2x+y=18$ …… (IV) By adding equation (III) and equation (IV): we get, \begin{align} & 5x-y=31 \\ & \underline{2x+y=18} \\ & 7x=49 \\ & \text{ }x=7 \\ \end{align} Putting in the value of $x=7$ in equation (III). We get, \begin{align} & 35-y=31 \\ & -y=-4 \\ & y=4 \end{align} Substitute $x=7$ and $y=4$ in equation (I): \begin{align} & 7+4+z=16 \\ & z=5 \end{align} Thus, the solution of the system of equations is $x=7,y=4\text{ and }z=5$.