## Precalculus (6th Edition) Blitzer

Let us consider the inequalities \begin{align} & {{x}^{2}}+{{y}^{2}}\le 16 \\ & x+y<2 \\ \end{align} Put the equals symbol in place of the inequality and rewrite the equation as given below: \begin{align} & {{x}^{2}}+{{y}^{2}}=16 \\ & x+y=2 \\ \end{align} The equation shows a circle with origin at $\left( 0,0 \right)$ of radius $4$. Second equation shows a line passing through the coordinates $\left( 0,2 \right)$. Now, take origin $\left( 0,0 \right)$ as a test point for the equations and check the region in the graph to shade: \begin{align} & {{x}^{2}}+{{y}^{2}}\le 16\text{ and }x\text{+}y<2 \\ & {{0}^{2}}+{{0}^{2}}\le 16\text{ and 0}<2 \\ & 0\le 16\text{ and 0}<2 \\ \end{align} As we see that the test point satisfies both inequalities, therefore the shaded region will contain the test point; that is, the origin in both cases. Thus, the graph of the provided inequality is plotted.