Question

which graph shows the solution to the system of linear inequalities?
x + 3y > 6
y ≥ 2x + 4

Explanation:

Step1: Analyze \(x + 3y>6\)

Rewrite it as \(y>-\frac{1}{3}x + 2\). The boundary line \(y = -\frac{1}{3}x+2\) has a slope of \(-\frac{1}{3}\) and a y - intercept of 2. Since the inequality is \(y>-\frac{1}{3}x + 2\), the line should be dashed (because the inequality is strict, \(>\) not \(\geq\)) and we shade above the line.

Step2: Analyze \(y\geq2x + 4\)

The boundary line \(y = 2x+4\) has a slope of 2 and a y - intercept of 4. Since the inequality is \(y\geq2x + 4\), the line should be solid (because of the \(\geq\) sign) and we shade above the line.

Step3: Match with graphs

• For the first inequality \(y>-\frac{1}{3}x + 2\), the dashed line with slope \(-\frac{1}{3}\) (going down from left to right) and shading above. For the second inequality \(y\geq2x + 4\), the solid line with slope 2 (going up from left to right) and shading above.
• Looking at the graphs, the third graph (the one with the dashed line \(y = -\frac{1}{3}x + 2\) and solid line \(y=2x + 4\) with the correct shading) matches. The dashed line has a slope of \(-\frac{1}{3}\) (dashed, since \(x + 3y>6\) is strict) and the solid line has a slope of 2 (solid, since \(y\geq2x + 4\) is non - strict). The shading for both inequalities (above each line) intersects in the region shown in the third graph.

Answer:

The third graph (the one with the dashed line \(y = -\frac{1}{3}x+2\) and solid line \(y = 2x + 4\) and the appropriate shading)