Exercise 39 Page 405

The given point $(\text{-}3,11)$ is a solution to a system of inequalities if it satisfies every inequality in the system. Let's substitute the point into the given systems one at a time until we find the correct one.

Option A

If the given point satisfies both inequalities at the same time, the point is a solution to the system. Let's substitute the point into both inequalities of the system. Because we obtained true statements from both inequalities by substituting the point $(\text{-}3,11),$ it is a solution to the system in option A.

Checking the Remaining Options

Although option A is correct, let’s verify our answer by checking the remaining options as well. To do so, we will following the same procedure for each of the remaining systems.

System	Substitution	Simplify
$$\{\begin{array}{l}y>x+8\\ 3x+y>2\end{array}$$	$$\{\begin{array}{l}11>\text{-}3+8\\ 3(\text{-}3)+11>2\end{array}$$	$$\{\begin{array}{l}11>5✓\\ 2>2\times \end{array}$$
$$\{\begin{array}{l}y>\text{-}x+8\\ 2x+3y\ge 7\end{array}$$	$$\{\begin{array}{l}11>\text{-}(\text{-}3)+8\\ 2(\text{-}3)+3(11)\ge 7\end{array}$$	$$\{\begin{array}{l}11>11\times \\ 27\ge 7✓\end{array}$$
$$\{\begin{array}{l}y\le \text{-}3x+1\\ x-y\ge \text{-}15\end{array}$$	$$\{\begin{array}{l}11\le \text{-}3(\text{-}3)+1\\ \text{-}3-11\ge \text{-}15\end{array}$$	$$\{\begin{array}{l}11\le 10\times \\ \text{-}14\ge \text{-}15✓\end{array}$$

When we substitute the point $(\text{-}3,11)$ into any of the remaining systems, at least one of the inequalities is not satisfied. Therefore, the point $(\text{-}3,11)$ is not a solution of any of the remaining systems. This means that, as we discovered at the beginning of our solution, the correct answer is indeed option A.