## Precalculus (6th Edition) Blitzer

The restaurant should should purchase $6$ two-seat tables and $11$ four-seat tables.
Let us assume x to be the number of two-seat tables and y to be the number of four-seat tables. The provided data can be represented in the form of equations as follows: $4x+2y=56$ (I) $x+y=17$ (II) Solve for x in the second equation to obtain its value in terms of y: \begin{align} & x+y=17 \\ & x=17-y \end{align} Put the value of $x$ from the second equation in the first equation: \begin{align} & 4x+2y=56 \\ & 4\cdot \left( 17-y \right)+2y=56 \end{align} Simplify: \begin{align} & 68-4y+2y=56 \\ & 68-56=2y \\ & 12=2y \end{align} And divide by $2$ on both sides \begin{align} & y=6 \\ & x=17-y \\ \end{align} Put the value $y=6$ in $x=17-y$: \begin{align} & x=17-6 \\ & x=11 \\ \end{align} Hence, the number of four-seat tables is $11$ and the number of two-seat tables is $6$.