Question

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.

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

The first line through points (-3, 0) and (-4, -1) is the orange line.

The second line, through (1, 1) and (2, -1) is the purple line.

Since they are both shaded below, the intersection of the shaded regions, is the region colored in blue.

Both lines are included, since they are solid lines.



The equation of the first line can be found as follows:

the slope is $\frac{0-(-1)}{-3-(-4)}=\frac{1}{1}=1$, thus, the equation is:
$y-0=1(x-(-4))\\\\y=x+4$


The equation of the second line can be found as follows:

the slope is $\frac{1-(-1)}{1-2}= \frac{2}{-1}=-2$, 

thus the equation of the second line is :


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


We have to decide on the direction of the inequality sign, so we pick a point, for example (0, 4) which is not in the colored regions of the lines, 

so we write the inequalities so that they do not hold for (0, 4)


line 1:

y≥x+4

4≥0+4=4 is true, so we change the sign and write : y≤x+4


similarly, line 2:

y≤-2x+3

4≤-2*0+3=3, which is not true, so we have the correct direction.


Note that since the lines were solid, we include the equality case:


Answer: 

the shaded region is the solution to the system of inequalities:

i) y≤x+4
ii) y≤-2x+3