# Graph the system of inequalities. y < 3/2x + 3 -y < 2x

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First, consider both inequalities as equalities by replacing < with =. Then we have `y=3/2x+3` and `-y=2x`. To graph, we need to solve the second equation for *y*, which means dividing both sides by -1 to yield `y=-2x`. We can now graph both lines, which represent boundary lines of the solution region. Because the inequalities are strictly less than, the lines must be dashed rather than solid.

To determine which regions to shade, we return to consideration of the original inequalities. For the first inequality (whose boundary is the black line), use a test point such as (0, 0) to see whether it makes the inequality true. Substituting 0 for both *x* and *y* gives us `0<3/2(0)+3`, which simplifies to `0<3`. Since the resulting inequality is true, we shade to the side of the black line that includes the test point (0, 0).

Similarly, choose a test point, such as (1, 1), for the second inequality. Then we get `-(1)<2(1)`, or `-1<2`, which is true, so the red line in shaded toward the point (1, 1). **The solution region for the system of inequalities is the overlap of the two shaded regions, which is the region to the right of the point of intersection.**

Graph the system of inequalities:

To graph the system you perform two steps:

(1) Graph the "lines"-- if the inequality is strict (<,>) then the line is dashed, otherwise (`<=,>=)` solid.

(2) Then you must determine the region to shade -- shading indicates every point in the region is a solution to the system.

Since there are two lines that intersect at a point, there will be 4 regions to choose from.

First graph the lines `y=3/2x+3` and `-y=2x` ; both lines will be dashed as no point on either line is a solution to the system. Here the graph of `y=3/2x+3` is in red, the other in black:

You can test a point from each region to determine the region to shade, or you can shade the graph for each inequality and then the answer is the overlapping shaded area.

**We find the point (1,1) satisfies both inequalities, so every point in the region containing (1,1) is a solution. (Shade "under" the black line and "above" the red line.)**

** Note that points in the other three regions will not satisfy both inequalities. e.g. (-1,-1) is not a solution to -y<2x since 1 is notless than -2, (-3,0) is not a solution to `y<3/2x+3` since 0 is not less than -1.5, and (0,4) is not a solution to `y<3/2x+3` since 4 is not less than 3. **