Respuesta :

kudzordzifrancis

Answer:

See graph

Step-by-step explanation:

The given system of inequalities is;

[tex]-2x+3y\ge-12[/tex]

[tex]x+y\:<\:3[/tex]

We graph [tex]-2x+3y\ge-12[/tex] by graphing the corresponding equation

[tex]-2x+3y=-12[/tex].

We can do this by plotting the intercepts (0,-4) and (6,0).

We draw a solid line through these points.

We test the inequality with the point (0,0).

[tex]-2(0)+3(0)\ge-12[/tex]

This is true so we shade the upper half plane.

We graph [tex]x+y\:<\:3[/tex] by also plotting the points (0,3) and (3,0) and draw a broken line through them.

We again test the origin

[tex]0+0\:<\:3[/tex]

This is true, so we shade the lower half plane.

The intersection of the two region is shown in the attachment.

Ver imagen kudzordzifrancis
Ver imagen kudzordzifrancis
Ver imagen kudzordzifrancis
zainebamir540

Answer:

Graph is attached. See it.

Step-by-step explanation:

We have given system of inequality :

-2x + 3y >= -12

x+y < 3

We have to plot a graph of this system of inequality.

For first inequality, we plot the associated equation -2x+3y = -12

Find the x and y intercepts and join them which the solid straight line.

then put (0,0) to check the inequality.

0<= -12

It satisfy the inequality. Now shade the upper half plane.

Now draw x+y = 3 by similar procedure. Draw (0,3) and (3,0) in the graph that are they and  x intercepts respectively.

Draw a dotted line joining these two points.

Testing the inequality, put (0,0) in the inequality x+y < 3 we get,

0<3.

It satisfy the inequality.

Shade the lower half plane.

The intersection of two inequalities are shown in the attached graph.

Ver imagen zainebamir540

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