Chapter 6.5, Problem 32PPE

To determine

To find: The linear inequality, represented by given graph.

Answer to Problem 32PPE

Required linear inequality is

  $3y+x\ge 1$.

Explanation of Solution

Given information: Below graph is given,

  

Concept used: Find the equation of line given using the below formula of equation of line passing from two points $(x_1,y_1)$ and $(x_2,y_2)$ is,

  $y-y_{`}{}_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

And then get its relevant inequality satisfied by $(0,0)$ to find its solution region.

Calculation: Shown line passes from points $(1,0)$ and $(0,3)$ , so its equation is,

  $\begin{array}{l}y-0=\frac{3-0}{0-1}(x-1)\\ \Rightarrow y=-\frac{1}{3}x+\frac{1}{3}\\ \Rightarrow 3y=-x+1\\ \Rightarrow 3y+x=1\end{array}$

And its equivalent linear inequality is,

  $3y+x\ge 1$

Check its shaded region by plug in $x=0,y=0$ in above inequality,

  $\begin{array}{l}3\times 0+0\ge 1\\ \Rightarrow 0\ge 1\end{array}$

That is false, so solution region is not towards point $(0,0)$ as shown in given graph.

Conclusion: So, given graph is of linear inequality $3y+x\ge 1$ .