Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions: y > −2x + 3 y is less than 1 over 2 times x minus 2 part a: describe the graph of the system, including shading and the types of lines graphed. provide a description of the solution area. (6 points) part b: is the point (−2, 3) included in the solution area for the system? justify your answer mathematically. (4 points)

(-2,3) is not a solution.

Step-by-step explanation:

Part A :

Let us first write the two equations

y>-2x+3

y<(1/2)x-2

In order to graph these inequalties , we first need to graph the equations

y=-2x+3

y=1/2 x-2

1) In first equation for

x=0 , y=3

x=1 , y= 1

Hence we can join coordionates (0,3) and (1,1) with a broken line. Now let us see whether (0,0) satisfies the inequation

0>-2(0)+3

0>0+3

0>3

Which is false , hence (0,0) does not lies in the solution , Hence shade the region which does not contain (0,0) . The region is the graph of inequation y>-2x+3

2) In second equation y=(1/2)x-2 for

x=0 ; y= -2

x=2 ; y=-1

Hence we join the coordinates (0,-2) and (2,-1) with broken lines. Now check whether (0,0) satisfies the inequation y<(1/2)x-2

0<(1/2)(0)-2

0<-2 which is not true , hence (0,0) does not lies in the graph of this inequation. Thus we shade the region which does not contain (0,0).

Now the shaded area which is common in these two inequations becomes our solution.

Part B:

please refere to the picture attached. here we first plot the coordinate (-2,3) and we see that this coordinate does not lies in the common shaded region. hence this is not a solution to the two inequations.

Also mathematically we can show that (-2,3) does not satisfy any inequation of these.

y>-2x+3

3>-2(-2)+3

3>4+3

3>7 ; Which is false

y<(1/2)x-2

3<(1/2)(-2)-2

3<-1-2

3<-3

Which is again false.

Hence this coordinate is not a solution to these pair of inequalities.

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step-by-step explanation:

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