Solve the system of two linear inequalities graphically. x>-1 y>=3 Find the region with points that satisfy both inequalities. Select the region you wish to be shaded: A, B,C, and D
To graph the system of inequalities, we first need to graph each individual inequality separately.
For the inequality x > -1, we draw a vertical line at x = -1. Since the inequality is greater than, we will shade the region to the right of the line.
Here is the graph for x > -1:
[asy]
draw((-3,0)--(3,0),black+linewidth(1.5),EndArrow);
draw((0,-3)--(0,3),black+linewidth(1.5),EndArrow);
draw((-1,-3)--(-1,3),red+dashed);
label("x = -1",(-1,-2.5),S);
draw((1,-3)--(1,3),red,linewidth(1.5));
dot((-1,0),red+linewidth(2));
label("$A$", (-3,0), SW);
label("$B$", (0, 0), S);
label("$C$", (3, 0), SE);
[/asy]
For the inequality y >= 3, we draw a horizontal line at y = 3. Since the inequality is greater than or equal to, we will shade the region above the line.
Here is the graph for y >= 3:
[asy]
draw((-3,0)--(3,0),black+linewidth(1.5),EndArrow);
draw((0,-3)--(0,3),black+linewidth(1.5),EndArrow);
draw((-3,3)--(3,3),red+dashed);
label("y = 3",(2.5,3),E);
draw((-3,1)--(3,1),red,linewidth(1.5));
dot((0,3),red+linewidth(2));
label("$D$", (0, 3), N);
[/asy]
To find the region with points that satisfy both inequalities, we need to find the overlapping region. This is the region that is shaded in both graphs. It is the region to the right of the vertical line x = -1 and above the horizontal line y = 3.
Here is the overlapping region:
[asy]
draw((-3,0)--(3,0),black+linewidth(1.5),EndArrow);
draw((0,-3)--(0,3),black+linewidth(1.5),EndArrow);
fill((1,3)--(1,1)--(-3,1)--(-3,3)--cycle,blue);
draw((-1,-3)--(-1,3),red+dashed);
label("x = -1",(-1,-2.5),S);
draw((1,3)--(1,-3),red,linewidth(1.5));
label("y = 3",(2.5,3),E);
label("$C$", (3, 0), SE);
label("$D$", (0, 3), N);
[/asy]
The shaded region is region D, which represents the points that satisfy both inequalities x > -1 and y >= 3.