Answer:
[tex]y\le -\frac{1}{3}x + 1[/tex]
Step-by-step explanation:
The equation of the line is y = mx + b
where
m is the slope = rise/run
b = y-intercept ie where the line crosses the y-axis at x = 0
The line crosses the y axis at y = 1 so the y-intercept is 1 and the point at which the line crosses is (0, 1)
We have the equation as
y = mx + 1
To find the slope, take any two points on the line. Find the corresponding difference in the y values and divide this difference by the corresponding difference in the x values
Two convenient points are (0, 1) and (3, 0)
Slope = [tex]\frac{0-1}{3-0} = -\frac{1}{3}[/tex]
Equation of the line :
[tex]y = -\frac{1}{3}x + 1[/tex]
To find out if the inequality in the shaded region is a ≥ or a ≤ inequality, take a point in the shaded region. Determine whether the chosen point x, y values satisfy which of the following equations
[tex]y\le -\frac{1}{3}x + 1[/tex] (1)
or
[tex]y\ge -\frac{1}{3}x + 1[/tex] (2)
The point (0,0) is inside the shaded region
Plug in these values into inequality (1)
[tex]0 = -\frac{1}{3}\cdot0 + 1[/tex] = 1
Is 0 ≤ 1 ?
Indeed it is so the inequality is
[tex]y\le -\frac{1}{3}x + 1[/tex] (Answer)
The region on the opposite side of the line must be a ≥ inqeuality
The attached graph makes it clearer
