How do you graph the system x - 4y >= -4 and 3x + y <= 6?

1 Answer
Feb 22, 2016

1) Graph the line y=1/4 x + 1,
it has a slope of 1/4 and a y intercept of 1.
2) The region x-4y>=-4 (or y<=1/4 x + 1) is the area below this line and the line itself, shade/hatch this region.

3) Graph the line y=-3x+6,
it has a slope of -3 and a y intercept of 6.
4) The region 3x+y<=6 (or y<=-3x+6) is the area below this line and the line itself, shade/hatch this region a different colour/pattern from the other region.

5) The SYSTEM, is the set of x and y values the satisfy both expressions. This is intersection of both regions. Whatever both shades occur is the graph of the system.

Explanation:

Consider the region defined by x-4y>=-4.
The edge of the region is defined by the equation x-4y=-4.
This need to be put in standard form.

Start with,
x-4y>=-4
Subtract x from both sides.
x-4y-x>=-4-x
Producing,
-4y>=-4-x.

Divide both side by -4 (remember to flip the inequality)
{-4y}/-4<={-4-x}/-4.
We have
y<=1+x/4 or y<=1/4 x + 1.
The edge is the line y=1/4 x + 1 and the region the area below this including the line.

Consider the region defined by 3x+y<=6.
The edge of the region is defined by the equation 3x+y=6.
This need to be put in standard form.

Start with,
3x+y<=6
Subtract 3x from both sides.
3x+y-3x<=6-3x
Producing,
y<=6-3x
or
y<=-3x+6

The edge is the line y=-3x+6 and the region the area below this including the line.

The SYSTEM, is the set of x and y values the satisfy both expressions. This is intersection of both regions.