Respuesta :

gmany

[tex]3y\geqx-9\qquad|:3\\\\y\geq\dfrac{1}{3}x-3\\\\3x+y > -3\qquad|-3x\\\\y > -3x-3[/tex]

[tex]\left\{\begin{array}{ccc}y\geq\dfrac{1}{3}x-3\\\\y > -3x-3\end{array}\right[/tex]


[tex]y=\dfrac{1}{3}x-3\\\\for\ x=0\to y=\dfrac{1}{3}(0)-3=-3\to(0,\ -3)\\\\for\ x=3\to y=\dfrac{1}{3}(3)-3=1-3=-2\to(3,\ -2)\\\\y=-3x-3\\\\for\ x=0\to y=-3(0)-3=-3\to(0,\ -3)\\\\for\ x=-1\to y=-3(-1)-3=3-3=0\to(-1,\ 0)[/tex]

[tex]y\geq\dfrac{1}{3}x-3[/tex]

solid line. we shade above the line

[tex]y > -3x-3[/tex]

the dotted line. shadow above the line

Answer in the attachment (the first graph).

Ver imagen gmany
danieltirado

Answer:

The first graph (picture of the graph attached)

Step-by-step explanation:

We have two inequalities:

3y ≥ x-9

3x+y > -3

We have to see which graph represents the solution to these inequalities.

We are going to solve for some values of x:

First inequality 3y ≥ x-9

y ≥ (x-9) / 3

If we give values to x then we solve for y:

x             y

-2            -3.67

0             -3

2              -2.34

4              1-67

With these values, we can graph the first line, which is continuous because the inequality has ≥

And because it is greater or equal to the shaded region is everything up from the line.

The second inequality 3x+y > -3

y > -3 - 3x

x            y

-4           9

-2           3

0            -3

Now we can graph the second inequality which will be a continuous line because it only has >

The shaded region has to be up from the line because it is greater than.

Ver imagen danieltirado

Otras preguntas