Question

Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions:

Answer 1

Answer:

(-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.

Answer 2

Y > 5x + 5 . . . (1)y > -1/2x + 1 . . . (2)
Part A: The graph of the system are two straight dotted lines with the first line passing through points (-1, 0) and (0, 5) with the region above the line shaded and the second line passing through (0, 1) and (2, 0) with the region above the line shaded. The two lines intersect at point (-8/11, 15/11).
Part B: The point (-2, 5) is included in the solution area.5 > 5(-2) + 5 and 5 > -1/2(2) + 15 > -10 + 5 and 5 > -1 + 15 > -5 and 5 > 0
hope i helped