Question

hi please help I have only 10 min to do this because it's due in 10 minutes and I've been trying to figure this out for a while.

Answer 1

`\begin{gathered} 3y>2x+12 \\ 2x+y\leq-5 \end{gathered}`

To graph a system of inequalities you need to find the coordinates of 2 points on each inequality (as they are lineal inequalities):

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`3y>2x+12`

Solve for y: divide both sides of the inequality into 3:

`\begin{gathered} (3y)/(3)>((2x+12))/(3) \\ \\ y>(2)/(3)x+(12)/(3) \\ \\ y>(2)/(3)x+4 \end{gathered}`

Find the points as in a equation:

If x is 0:

`\begin{gathered} y>(2)/(3)(0)+4 \\ y>0+4 \\ y>4 \end{gathered}`

First point (0 , 4)

If x is 6

`\begin{gathered} y>(2)/(3)(6)+4 \\ y>(12)/(3)+4 \\ y>(12+12)/(3) \\ y>(24)/(3) \\ y>8 \end{gathered}`

Second point ( 6, 8 )

As the inequality is y greather than (2/3x+4) you draw a dot line that go through the two points and shade the area over the line:

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Repeat the process to find two points in the second inequality:

`\begin{gathered} 2x+y\leq-5 \\ \\ y\leq-2x-5 \end{gathered}`

If x is 0:

`\begin{gathered} y\leq-2(0)-5 \\ y\leq0-5 \\ y\leq-5 \end{gathered}`

First point (0 , -5)

If x is -5

`\begin{gathered} y\leq-2(-5)-5 \\ y\leq10-5 \\ y\leq5 \end{gathered}`

Second point ( -5 , 5)

As the inequality is y less than or equal to (-2x-5) you draw a line that go through the two points and shade the area under that line:

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Then, you get the next graph for the system of inequalities:

The solution is the area shaded by both inequalities.