Graph the system of inequalities x + y ≤ 6 and x ≥ 2, and describe the solution region.

Step-by-step solution

1. 1

   Rewrite the first inequality with $y$ isolated. Subtracting $x$,

   $y\le 6-x.$

   Putting it in "$y\le$" form tells you immediately which side of the boundary line to shade: below it, not above.

2. 2

   Draw the two boundary lines. The boundary of the first is $y=6-x$, a line of slope $-1$ through $(0,6)$ and $(6,0)$. The boundary of the second is the vertical line $x=2$. Both inequalities are non-strict ($\le$, $\ge$), so both lines are drawn solid and belong to the solution.

3. 3

   Shade each region and take the overlap. For $y\le 6-x$, shade everything on or below the slanted line; for $x\ge 2$, shade everything on or to the right of $x=2$. The solution set is the intersection of the two shaded half-planes.

4. 4

   Locate the corner. The two boundaries meet where $x=2$ and $y=6-2=4$, so the vertex of the region is $(2,4)$. From there the region opens downward and to the right — it is unbounded, since nothing prevents $x$ or $-y$ from growing.

5. 5

   Write the region and test points. The solution is

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

   Check $(3,1)$: $3\ge2$ ✓ and $3+1=4\le6$ ✓, so it is in the region. Check $(1,1)$: $1\ge2$ fails, so it is out. Check $(4,3)$: $4\ge2$ ✓ but $4+3=7>6$, so it is out. The corner $(2,4)$ itself satisfies both with equality and is included.