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The graph below represents which system of inequalities? graph of two infinite lines that intersect at a point. One line is solid and goes through the points negative 3, 0, negative 4, negative 1 and is shaded in below the line. The other line is solid, and goes through the points 1, 1, 2, negative 1 and is shaded in below the line. y ≤ −2x + 3 y ≤ x + 3 y ≥ −2x + 3 y ≥ x + 3 y ≤ −3x + 2 y ≤ −x + 2 y > −2x + 3 y > x + 3

Respuesta :

texaschic101
solid...(-3,0)(-4,-1)...shaded below the line

slope = (-1-0) / (-4-(-3) = -1/(-4 + 3) = -1/-1 = 1

y = mx + b
-1 = 1(-4) + b
-1 = -4 + b
-1 + 4 = b
3 = b

this line is : y < = x + 3 <== (thats less then or equal)
====================
solid...(1,1)(2,-1)...shaded below the line

slope = (-1-1) / (2-1) = -2/1 = -2

y = mx + b
1 = -2(1) + b
1 = -2 + b
1 + 2 = b
3 = b

this line is : y < = -2x + 3 <== (thats less then or equal)

The system of inequalities are: [tex]y\leq x+3[/tex]  and  [tex]y\leq -2x+3[/tex] and this can be determined by using the point-slope form of the line.

Given :

  • One line is solid and goes through the points (-3,0) and (-4,-1) and is shaded below the line.
  • The other line is solid and goes through points (1,1) and (2,-1) and is shaded below the line.

The equation of the line passing through (-3,0) and (-4,-1) is:

[tex]\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

[tex]\dfrac{y-0}{x+3}=\dfrac{-1-0}{-4+3}[/tex]

[tex]y(-1)=-(x+3)[/tex]

[tex]y=x+3[/tex]

Given that the shaded region is below the line, that is:

[tex]y\leq x+3[/tex]

The equation of the line passing through (1,1) and (2,-1) is:

[tex]\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

[tex]\dfrac{y-1}{x-1}=\dfrac{-1-1}{2-1}[/tex]

[tex]y-1= -2(x-1)[/tex]

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

Given that the shaded region is below the line, that is:

[tex]y\leq -2x+3[/tex]

For more information, refer to the link given below:

https://brainly.com/question/19110185

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