Chapter 6.6, Problem 45PFA

To determine

To find: System of inequalities of given graph.

Answer to Problem 45PFA

  $\begin{array}{l}-x+y\ge 3\\ 2x+y\le 5\end{array}$

Explanation of Solution

The left bold line passes through the points $\left(-3,0\right)$ and $\left(0,3\right)$

So, slope of the line is

  $\begin{array}{l}{m}_1=\frac{y_2-y_1}{x_2-x_1}\\ \Rightarrow m_1=\frac{3-0}{0-\left(-3\right)}\\ \Rightarrow m_1=\frac{3}{3}\\ \Rightarrow m_1=1\end{array}$

From point slope form of line

  $\begin{array}{l}\left(y-y_1\right)=m\left(x-x_1\right)\\ \Rightarrow y-0=1\left(x-\left(-3\right)\right)\\ \Rightarrow y=x+3\\ \Rightarrow -x+y=3\end{array}$

It is a solid line and shaded region lies above that line,

So, the sign of inequality must be “ $\ge$ ”

Therefore, one of the inequality is $-x+y\ge 3$

The right bold line passes through the points $\left(2,1\right)$ and $\left(1,3\right)$ .

So, slope of the line is

  $\begin{array}{l}{m}_2=\frac{y_2-y_1}{x_2-x_1}\\ \Rightarrow m_2=\frac{3-1}{1-2}\\ \Rightarrow m_2=\frac{2}{-1}\\ \Rightarrow m_2=-2\end{array}$

From point slope form of line

  $\begin{array}{l}\left(y-y_1\right)=m_2\left(x-x_1\right)\\ \Rightarrow \left(y-1\right)=-2\left(x-2\right)\\ \Rightarrow y-1=-2x+4\\ \Rightarrow y+2x=1+4\\ \Rightarrow 2x+y=5\end{array}$

It is a solid line and shaded region lies below that line,

So, the sign of inequality must be “ $\le$ ”

Therefore, one of the inequality is $2x+y\le 5$ .

Therefore, the system of inequalities are

  $\begin{array}{l}-x+y\ge 3\\ 2x+y\le 5\end{array}$