Question

solve the system of two linear inequalities graphically. {7x + 6y < 42, x ≥ 3}
step 2 of 3: graph the solution set of the second linear inequality.
choose the type of boundary line:
solid (—)
dashed (--)
enter two points on the boundary line: ( , )( , )
select the region you wish to be shaded:

Explanation:

Step1: Analyze the first inequality $x\geq3$

The boundary line is $x = 3$. Since the inequality is $\geq$, the boundary line is solid. Two points on the line $x = 3$ can be $(3,0)$ and $(3,5)$.

Step2: Analyze the second - inequality $7x+6y<42$

Rewrite it in slope - intercept form $y=mx + b$. First, solve for $y$: $6y<-7x + 42$, so $y<-\frac{7}{6}x+7$. The boundary line is $y =-\frac{7}{6}x + 7$. Since the inequality is $<$, the boundary line is dashed. To find two points on the line $y=-\frac{7}{6}x + 7$, when $x = 0$, $y = 7$ (the point $(0,7)$); when $y = 0$, $7x=42$, $x = 6$ (the point $(6,0)$).

Step3: Determine the shaded region

For $x\geq3$, we shade to the right of the line $x = 3$. For $y<-\frac{7}{6}x+7$, we test a point not on the line, say $(0,0)$. Substitute into $y<-\frac{7}{6}x + 7$: $0<7$ (true), so we shade the region below the line $y=-\frac{7}{6}x + 7$. The intersection of the two regions (right of $x = 3$ and below $y=-\frac{7}{6}x + 7$) is the solution set.

The type of boundary line for $x\geq3$: Solid. Two points on $x = 3$ are $(3,0)$ and $(3,5)$.
The type of boundary line for $7x+6y<42$: Dashed. Two points on $y=-\frac{7}{6}x + 7$ are $(0,7)$ and $(6,0)$.

Answer:

For $x\geq3$: Solid, points $(3,0),(3,5)$
For $7x + 6y<42$: Dashed, points $(0,7),(6,0)$
The shaded region is to the right of $x = 3$ and below $y=-\frac{7}{6}x + 7$.