Answer
See below:
Work Step by Step
Let one variable be x and another variable be y.
Sum of two variables is at most 3, which gives
\[x+y\le 3\](1)
The sum ofvariable y with 4 times the variable x does not exceed 6, which gives
\[4x+y\le 6\](2)
Both the inequalities in equation (1) and equation (2) are of the same sign.
So, treat the equation as equality problem, which gives
\[x+y=3\] (3)
\[4x+y=6\] (4)
Subtract equation (3) from (4) by elimination process to get
\[\begin{align}
& 3x=3 \\
& x=1
\end{align}\]
Put the value of \[x=1\]in\[x+y=3\], which gives
\[\begin{align}
& 1+y=3 \\
& y=2
\end{align}\]
Both the equations intersect at\[\left( 1,2 \right)\].
Put \[y=0\], which gives the x-intercept.
So, the x-intercept for equation (3) is
\[\begin{align}
& x+y=3 \\
& x+0=3 \\
& x=3
\end{align}\]
Put \[x=0\], which gives the y-intercept.
The y-intercept for equation (3) is
\[\begin{align}
& x+y=3 \\
& 0+y=3 \\
& y=3
\end{align}\]
Similarly, the x-intercept for equation (4) is
\[\begin{align}
& 4x+y=6 \\
& 4x+0=6 \\
& x=\frac{6}{4} \\
& x=1.5
\end{align}\]
The y-intercept for equation (4) is
\[\begin{align}
& 4x+y=6 \\
& 4\left( 0 \right)+y=6 \\
& 0+y=6 \\
& y=6
\end{align}\]
Now, take test point for equation (1):
Let test point be \[\left( 0,0 \right)\].
Put it in equation (1), which gives
\[\begin{align}
& x+y\le 3 \\
& 0+0\le 3 \\
& 0\le 3
\end{align}\]
The above statement is true. The shaded region contains the test point\[\left( 0,0 \right)\].
Hence, the shaded region is given as
