Question

What system of linear inequalities is shown in the graph? Enter your answers in the boxes.

Answer 1

Answer:

y ≤ (-2)x+2

y < (1/3)x + 2

Step-by-step explanation:

First, we'll want to figure out the lines that are intersecting.

To find the equation of a line given two points (of form y=mx+b), we can first find the slope (m). This is equal to (y₂-y₁)/(x₂-x₁) for points (x₁,y₁) and (x₂,y₂). Then, we can plug a point in and solve for b.

For the line that has points at (1,0) and (2, -2), our slope is

(y₂-y₁)/(x₂-x₁) = (-2-0)/(2-1)

= (-2)/1

= -2

Our equation is therefore y= (-3/2)x+b. Plugging a point like (2,-2) in, we get

-2 = (-2)(2) + b

-2 = -4 + b

add 4 to both sides to isolate b

b =2

Our equation is thus y = (-2)x+2

For the other line, with points at (-3,1) and (0,2), our slope is (2-1)/(0-(-3)) = 1/3

Then, our equation is y=(1/3)x+b

plug (-3,1) in

1 = (1/3)(-3) + b

1 = -1 + b

add 1 to both sides to isolate the b

b =2

Our equation is thus y=(1/3)x + 2

Our equations are y=(1/3)x + 2 and y = (-2)x+2. In the graph shown, the shaded area is under both the lines (it is also to the left of them, not to the right or on top). Therefore, we can say that the shaded area is less than the line.

For y = (-2)x+2, the line is not dotted, so the line is included in the region. This means that our inequality is y ≤ (-2)x+2 (note the ≤ rather than simply the < sign)

For y=(1/3)x + 2, the line is dotted, so it is not included in the region. This means that our inequality is y < (1/3)x + 2