Answer
See below:
Work Step by Step
Let one variable be x and another variable be y.
Sum of two variablesis at most 4.It gives
\[x+y\le 4\] (1)
The sum ofvariable y with 3 times the variable x does not exceed 6.It gives
\[3x+y\le 6\] (2)
Since both the inequalities in equation (1) and equation (2) are of the same sign, treat the equations as equality problem.It gives
\[\begin{align}
& 3x+y=6 \\
& x+y=4
\end{align}\]
Subtract equation (1) from (2).
\[\begin{align}
& 2x=2 \\
& x=1
\end{align}\]
Put the value of \[x=1\]in\[x+y=4\]. It gives
\[\begin{align}
& y+1=4 \\
& y=3
\end{align}\]
Both the equations intersect at\[\left( 1,2 \right)\].
Now, we need to find the x-intercept and y-intercept of both the equations.
\[3x+y=6\]
For x-intercept,\[y=0\].
\[\begin{align}
& 3x+y=6 \\
& 3x=6 \\
& x=2 \\
\end{align}\]
So, the x-intercept of equation (1) is\[(2,0)\].
For y-intercept, \[x=0\].
\[\begin{align}
& 3(0)+y=6 \\
& y=6 \\
\end{align}\]
So, the y-intercept of the equation (1)is\[(0,6)\].
Again, for x-intercept, \[y=0\].
\[\begin{align}
& x+y=4 \\
& x=4 \\
\end{align}\]
So, the x-intercept of the equation (2) is\[(4,0)\].
For y-intercept, \[x=0\].
\[\begin{align}
& x+y=4 \\
& y=4 \\
\end{align}\]
So, the y-intercept of the equation (2) is\[(0,4)\].
The graph is given as
