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For graphing systems of inequalities we use a Cartesian plane or XY -plane. In that vertical line divides the plane in left and right part and slanting or oblique line divides the plane in upper and lower part. A point in the Cartesian plane will either lie on a line or will lie in either of the part.

(i) Solve the given inequality for y by following the rules of inequality of signs.

(ii) Graph the equation either using function table or using slope and y-intercept.

(iii) If the inequality is strict (< or >), graph a dotted line and if the inequality is not strict ($\geq or \leq$ ), draw a solid line.

(iv) Apply the

Origin Test : Put x = 0 and y = 0 in the given equation after solving if the equation is true then origin (0,0) is included in the region and if false then set the region in which the origin is not included. |

Lightly shade the half-plane that is the graph of the inequality. Colored pencils may help you distinguish the different half-planes Graph both the equations using the above rules, mark the region accordingly.For marking use the two different color pencils or dark and light shade to distinguish the two different region.The intersection or overlapping region formed by both the inequalities is the solution region.

and y < x + 3 ------- (2)

Graph the 1st and 2nd equation using slope and y intercept.

Now put x = 0 and y = 0 in the 1st equation ,

0 $\geq$ - 3(0) -2

0 $\geq$ -2

Which is true, so origin is included for the 1st inequality.

So after graphing an equation we will mark right region of the line which includes origin.

2nd equation ⇒

0 < 0 + 3

0 < 3

which is also true, so origin is also included in 2nd inequality.

So after graphing an equation we will mark the lower region of the line which includes origin.

From the above graphs we can see that in A region from 1st graph, there is a solid line as the inequality is not strict. In region B, 2nd graph there is a dotted line as there is strict inequality. The 3rd graph both region A and B are overlapped and region c is intersection region. So region C is the solution.

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