Describe and graph the region of points satisfying both x + y <= 6 and x >= 2.

Step-by-step solution

1. 1

   Treat the two inequalities as one system. A point belongs to the solution only if it satisfies both, so the answer is the intersection of two half-planes — not their union:

   $\begin{cases}x+y\le 6\\ x\ge 2\end{cases}$

2. 2

   Put the first inequality in slope-intercept form. Subtracting $x$ from both sides:

   $y\le 6-x$

   The boundary $y=6-x$ is a line of slope $-1$ through $(0,6)$ and $(6,0)$, drawn solid because of the $\le$. Shade everything on or below it.

3. 3

   Interpret the second inequality. $x\ge 2$ is a vertical boundary line $x=2$, again solid, with everything on or to the right of it shaded. Note that no $y$ appears, so this restriction is the same at every height.

4. 4

   Locate the corner. The two boundaries meet where $x=2$ and $y=6-x$, i.e. at

   $(2,\,4)$

   This is the only vertex of the region; it belongs to the solution because both boundaries are included.

5. 5

   Check the shape is unbounded. Starting at $(2,4)$ the region opens downward and to the right without limit: for instance $(100,-1000)$ satisfies $x\ge 2$ and $x+y=-900\le 6$. So the region is an infinite wedge, not a triangle — there is no third boundary to close it off.

6. 6

   State the region. Test $(3,0)$: $3\ge 2$ $\checkmark$ and $3+0=3\le 6$ $\checkmark$, so it is inside. The solution set is

   $\{(x,y): x\ge 2\ \text{and}\ y\le 6-x\}$