Question

Graph the system of inequalities {y > 3x+2 and y<-2x+1. Which two quadrants does the solution lie in?

Answer 1

Quadrants 2 and 3

Explanation

$\begin{cases}y>3x+2 \\ y\leq-2x+1\end{cases}$

Step 1

graph inequality 1

$y>3x+2$

a)Plot the "y=" line (make it a solid line for y≤ or y≥, and a dashed line for y< or y>)

so

$\begin{gathered} y=3x+2 \\ i)\text{ for x= 0} \\ y=3(0)+2=0+2=2 \\ so \\ A(0,2) \\ i)\text{ for x= -1} \\ y=3(-1)+2=-3+2=-1 \\ B(-1,-1) \end{gathered}$

draw a line that passes trough P1 ( 0,2) and P2( -1,-1) and

b)shade below the line for a "less than" (y< or y≤).

so

$y>3x+2$

Step 2

graph inequality 2

$\begin{gathered} y\leq-2x+1 \\ \end{gathered}$

a)Plot the "y=" line (make it a solid line for y≤ or y≥, and a dashed line for y< or y>)

so

$\begin{gathered} y=-2x+1 \\ i)\text{ for x= 0} \\ y=-2\cdot(0)+1=0+1=1 \\ so \\ C(0,1) \\ i)\text{ for x= -1} \\ y=-2(-1)+1=2+1=3 \\ D(-1,3) \end{gathered}$

draw a line that passes trough C ( 0,1) and D( -1,3) and

b)shade below the line for a "less than" (y< or y≤).

so

$\begin{gathered} y\leq-2x+1 \\ \end{gathered}$

Step 3

finally, the solution is the intersection of the shaded areas,hence

therefores, the solution lies in

Quadrants 2 and 3

I hope this helps you