Q. 53 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 53–56, write a system of linear inequalities for the given graph.

Step-by-Step Solution

Answer

The solution is

$\left\{\begin{array}{l}x+y\le 6\\ x\le 4\\ x\ge 0\\ y\ge 0\end{array}\right.$

Step 1. Given information

The graph is in the first quadrant so the values are $x\ge 0$&$y\ge 0$.

Step 2. Given points

The given points of the line are $(0,6)$ & (4,2).

slope $$\begin{gathered} m=(\raisebox{1ex}{$y_2-y_1$}\!\left/ \!\raisebox{-1ex}{$(x_2-x1)$}\right. \\ m=\raisebox{1ex}{$(2-6)$}\!\left/ \!\raisebox{-1ex}{$($}\right.4-0) \\ m=\raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$4$}\right. \\ m=-1 \end{gathered}$$

Step 3. Line equation

$$\begin{gathered} y-y_1=m(x-x_1) \\ y-0=-1(x-6) \\ y=-x+6 \\ x+y=6 \end{gathered}$$

Since the origin lies in the graph so the function is $x+y\le 6$.

Step 4. Llast line

The line parallel to y-axis equation is $x=k$

here $k=4$ so the equation is $x=4$

The point origin lies in the graph so inequality is $x\le 4$.

Step 5. Solution

By considering all equations the inequalities are

$\left\{\begin{array}{l}x+y\le 6\\ x\le 4\\ x\ge 0\\ y\ge 0\end{array}\right.$