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# How do you know which way to shade a system of inequalities?

my problem is:
y (less than or equal to) (-2/3)X + 3

y (less than or equal to) (4/3)X - 3

I know how to graph it I just do not understand the shading and how to know which way to shade it, if you could help me it'd be appreciated!

Simply put, test a couple of numbers.

Once you've drawn the line on your graph (in simple problems) you will have two areas. Use an x value that fits inside one of the shaded areas and calculate the corresponding y value.

If that x value satisfies the inequality shade in that area of the in equality
When you are looking at a problem like this, think of the y-axis as your dividing line. Because both of these equations are linear, you will have to shade either above or below.

Start with the first equation and look at any number for x and plug it into your equation. You will solve for y (This will give you an x and y coordinate. This can be done several times, and each time, the resulting  x and y numbers should be in the area you want to shade (ie: below the line because y in this case is less than) Because there is an equal sign underneath, the line will be solid.

For the second equation, it also shows less than or equal to (again plug in numbers as above for x and solve for y to get your points) Again this line is solid (hence the equal sign)

To understand where the main area of shading is, gets a bit tricky, but if you did it right, you will have and area where the shaded regions of the individual graphs meet. This is where your final shading should be!

Hope this helps!
Graph the two lines.
Pick four points in each of the four quadrants around the intersection.

Point of intersection is -2/3x + 3 = 4/3x-3 => 3 + 3 = 4/3x+ 2/3x => 6 = 6/3x
x = 3

Use either equations and plug in x = 3 to find y.
y = 4/3(3) - 3 = 1

Pick x=0,y=0 plug them into each equations and see which is true.

1 <= -2/3(3)+3 => 1 <= -2+3
1 <= 1 true.

Point 0,0 is less that or equal to that equation. This side of equation one is shaded (not all of it. check the second equation).