Q. Which graph shows the solution to the system of linear inequalities? \(x + 3y > 6\) and \(y \geq 2x + 4\)

Problem: Solve the system of inequalities
\[
x+3y>6,\qquad y\ge 2x+4
\]

Step 1 — First inequality

Treat the boundary as an equation:
\[
x+3y=6.
\]
Solve for y to get slope-intercept form:
\[
3y=-x+6\quad\Rightarrow\quad y=-\tfrac{1}{3}x+2.
\]
This line has y-intercept \((0,2)\) and slope \(-\tfrac{1}{3}\). Because the inequality is strict (\(>\)), the boundary is dashed. Test the origin to determine the side to shade:
\[
0+3(0)>6 \quad\text{gives}\quad 0>6,\ \text{false}.
\]
Thus shade the half-plane of points satisfying
\[
y>-\tfrac{1}{3}x+2,
\]
i.e. the region above the dashed line.

Step 2 — Second inequality

Boundary line:
\[
y=2x+4.
\]
This line has y-intercept \((0,4)\) and slope \(2\). Because the inequality is nonstrict (\(\ge\)), draw this boundary as a solid line. Test the origin:
\[
0\ge 2(0)+4 \quad\text{gives}\quad 0\ge 4,\ \text{false}.
\]
So shade the half-plane of points satisfying
\[
y\ge 2x+4,
\]
i.e. the region above the solid line.

Step 3 — Intersection of the boundary lines

Find the intersection by substituting \(y=2x+4\) into \(x+3y=6\):
\[
x+3(2x+4)=6
\]
\[
x+6x+12=6
\]
\[
7x=-6\quad\Rightarrow\quad x=-\tfrac{6}{7}.
\]
Then
\[
y=2\Bigl(-\tfrac{6}{7}\Bigr)+4=-\tfrac{12}{7}+\tfrac{28}{7}=\tfrac{16}{7}.
\]
So the lines intersect at
\[
\Bigl(-\tfrac{6}{7},\tfrac{16}{7}\Bigr)\approx(-0.857,\,2.286).
\]

Final description of the solution set

The solution region is the overlap of the two half-planes:
\[
\{(x,y)\mid y>-\tfrac{1}{3}x+2\ \text{and}\ y\ge 2x+4\}.
\]
On the graph this appears as the wedge-shaped region above both lines, bounded below by the dashed line \(y=-\tfrac{1}{3}x+2\) and the solid line \(y=2x+4\), meeting at \(\bigl(-\tfrac{6}{7},\tfrac{16}{7}\bigr)\).

Key points for sketching:

• Dashed line through \((0,2)\) and \((6,0)\).
• Solid line through \((0,4)\) and \((-2,0)\).
• Shaded region: intersection of the half-planes above these two lines, starting at \(\bigl(-\tfrac{6}{7},\tfrac{16}{7}\bigr)\).