# On the graph solve the following system of inequalities. y<2x+3 `y>=-x`Check the answer by substituting back into the original...

On the graph solve the following system of inequalities. y<2x+3 `y>=-x`

Check the answer by substituting back into the original inequalities.

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Solve the system of inequlaities graphically:

`y<2x+3;y>=-x`

We graph the two inequalities in the same coordinate system. We graph as if they were equalities (lines) except (a) if the inequality is not inclusive (y<a) then we use a dotted line; if the inequality is inclusive (`y>=a` ) then we use a solid line and (b) the inequality divides the plane into half-planes; we shade the half-plane that includes the solution points.

The intersecting lines will cut the plane into 4 parts: 1 part will not have solutions to either inequality; 2 parts will have solutions to one inequlaity only; the last part will have solutions to both inequalities.

The graph:

For `y<2x+3` (the red line) notice we used a dotted line.In order to determine which half-plane to shade we pick a point **not** on the line. In this case (0,0) will work. Check to see if (0,0) is a solution point: Is 0<2(0)+3? Yes -- so every point on the same half-plane as (0,0) is a solution. We shade "underneath" or to the "right" of the red dotted line. ** If you had chosen say (-2,2), you would find that `2<2(-2)+3` was incorrect -- so no point on that half-plane is a solution so you shade the other half-plane**

For `y>=-x` we used a solid line. Again we check a point not on the line -- say (1,1). Then `1>=-(1)` is true, so we shade "above" the line.

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**The area that includes all solutions to both constraints (inequalities) is "below" the red dotted line and "above" the solid line.**

The grapher in this program will not allow shading.

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For part B, select a point in this region -- say (1,1). We check (1,1) in both inequalities:

1<2(1)+3 is true and `1>=-(1)` is also true.