Question

(brainlesest) Which system of inequalities does the graph represent? Which test point satisfies both of the inequalities in that system?
The graph represents the system of inequalities__________

A) 2x + 3y is greater than or equal to 4 and x+2y is greater than or equal to 3

B) 2x + 3y is less than or equal to 4 and x + 2y is less than or equal to 3

C) 2x + 3y is greater than or equal to 4 and x + 2y is less than or equal to 3

D) 2x +3y is less than or equal to 4 and x + 2y is greater than or equal to 3

E) 2x + 3y is less than or equal to 4 and 2x + 2y is less than or equal to to 3

The test point______satisfies both of the inequalities in the system represented by the graph.

A) (-3,3)
B) (0,2)
C) (2,1)
D) (4,-1)

Answer 1

Answer:

The answer is C for first question.

Only D satisfies both of the inequalities in the system represented by the graph. (Please check the graph attached)

Step-by-step explanation:

Lets say that red inequality graph is Graph A, blue inequality graph is Graph B.

First need to find slope of Graph A. Take two points from the graph. (-1,2) and  (3,0)

$m= (y_{2}-y_{1})/(x_{2}-x_{1})= (0-2)/(3-(-1)) =-2/4=-1/2$)

Slope of Graph A is -1/2. Lets do it for Graph B. Take two points from Graph B also (-1,2) and (2,0)

$m= (y_{2}-y_{1})/(x_{2}-x_{1})= (0-2)/(2-(-1)) =-2/3$)

Than lets find the in equation with using this slope:

$y-2=(-1/2)*(x-(-1))\\y-2=(-x/2)-1/2\\y=3/2-x/2$

Below side of the inequality is darked so the inequality A is:

$y$

Lets find the other inequality:

$y-2=(-2/3)*(x-(-1))\\y-2=(-2x/3)-2/3\\y=(-2x/3)+4/3$

Upper side of the inequality is darked so the inequality B is:

$y=>(-2x/3)+4/3$

Correct answer is C

Answer 2

Answer:Which system of inequalities does the graph represent? 

Which test point satisfies both of the inequalities in that system?

The graph represents the system of inequalities 

2x + 3y is less than or equal to 4 and x + 2y is greater than or equal to 3 

The test point (4, -1)satisfies both of the inequalities in the system represented by the graph