1. 6.5 Slope intercept form for Inequalities:
Linear Inequality: is a linear equation with an inequality sign (< , ≤, >, ≥)
Solution of an Inequality: is an ordered pair (x, y) that makes the inequality true.

2. GOAL:

3. Whenever we are given a graph we must be able to provide the equation of the function.
Slope-Intercept Form: The linear equation of a nonvertical line with an inequality sign:
y (<, ≤, >, ≥) m x + b
Slope = 𝒚𝟐−𝒚𝟏 𝒙𝟐−𝒙𝟏 = 𝑹𝒊𝒔𝒆 𝑹𝒖𝒏 = ⍙𝒚 ⍙𝒙
y-intercept y crossing

4. y < m x + b dash line shade left or down
Whenever we are given a graph we must be able to provide the equation of the function.
dash line shade left or down
y < m x + b

5. y > m x + b dash line shade right or up
Whenever we are given a graph we must be able to provide the equation of the function.
dash line shade right or up
y > m x + b

6. y ≤ m x + b Solid line shade left or down
Whenever we are given a graph we must be able to provide the equation of the function.
Solid line shade left or down
y ≤ m x + b

7. y ≥ m x + b Solid line shade Right or up
Whenever we are given a graph we must be able to provide the equation of the function.
Solid line shade Right or up
y ≥ m x + b

8. EX: Provide the equation of the inequality.

9. 2. Find another point to get the slope.
Solution: Since line is dashed and shaded at the bottom we use <. Also, the inequality must be in slope-intercept form:
Y < mx + b
A(0,1)
1. Find the y-intercept
B(3,-2)
In this graph b = +1.
2. Find another point to get the slope.
A(0,5) B(3,-2)

10. Use the equation of slope to find the slope:
𝒚𝟐−𝒚𝟏 𝒙𝟐−𝒙𝟏 = −𝟐−𝟏 𝟑−𝟎 = −𝟑 𝟑 = -1
A(0,1)
The slope-intercept form inequality is: y < -1x + 1
B(3,-2)
Remember: This means that if you start a 1 and move down one and over to the right one, and continue this pattern. We shade the bottom since it is <.

11. When work does not need to be shown: (EOC Test)
look at the triangle made by the two points.
Count the number of square going up or down and to the right.
In this case 1 down and 1 right. Thus slope is -1/1 = -1

12. YOU TRY IT Provide the equation of the inequality.

13. YOU TRY IT: (Solution) The inequality is solid and shaded below:
Y ≤ mx + b
1. Find the y-intercept
A(0,4)
In this graph b = + 4.
B(1,0)
2. Find another point to get the slope.
A(0,4) B(1,0)

14. Use the equation of slope to find the slope:
𝒚𝟐−𝒚𝟏 𝒙𝟐−𝒙𝟏 = 𝟎−𝟒 𝟏−𝟎 = −𝟒 𝟏 = - 4
A(0,4)
The slope-intercept form equation is: y ≤ -4x + 4
B(1,0)
Remember: This means that if you start a 4 and move down four and one over to the right. Solid line and shaded down means we must use ≤.

15. When no work is required, you can use the rise/run of a right triangle between the two points:
Look at the triangle, down 4 (-4) over to the right 1 (+1) slope = -4/+1 = -4
B(1,0)
Remember: You MUST KNOW BOTH procedures, the slope formula and the triangle.

16. Given Two Points: We can also create an inequality in the slope-intercept form from any two points and the words: less than (<), less than or equal to (≤), greater than(>), greater than and equal to(≥) accordingly.
EX: Write the slope-intercept form of the line that is greater than or equal to and inequality that passes through the points
(0, -0.5) and(2, -5.5)

17. Use the given points and equation of slope:
B(2,-5.5)
𝒚𝟐−𝒚𝟏 𝒙𝟐−𝒙𝟏 = −𝟓.𝟓−−𝟎.𝟓 𝟐−𝟎 = −𝟓 𝟐 = - 𝟓 𝟐
We now use the slope and a point to find the y intercept (b).
y ≥ mx + b
-3 = - 𝟓 𝟐 (𝟏) + b
-3 + 𝟓 𝟐 = b
Isolate b:
b = - 𝟔 𝟐 + 𝟓 𝟐 = - 𝟏 𝟐

18. y ≥ − 𝟓 𝟐 x – ½ y = mx + b m = - 𝟓 𝟐 and b = - ½
Going back to the equation:
y = mx + b
we replace what we have found:
m = - 𝟓 𝟐 and b = - ½
To get the final slope-intercept form of the line passing through (3, -2) and(1, -3)
y ≥ − 𝟓 𝟐 x – ½

19. Y-intercept y crossing
We now proceed to graph the equation:
y ≥ - 𝟓 𝟐 x - 𝟏 𝟐
Y-intercept y crossing
𝑹𝒊𝒔𝒆 𝑹𝒖𝒏

20. YOU TRY IT: Write the equation of the inequality.

21. y < mx + b A(3,-2) B(1,-3) 𝒚𝟐−𝒚𝟏 𝒙𝟐−𝒙𝟏 = −𝟑−−𝟐 𝟏−𝟑 = −𝟏 −𝟐 = 𝟏 𝟐
Use the given points and equation of slope:
A(3,-2)
B(1,-3)
𝒚𝟐−𝒚𝟏 𝒙𝟐−𝒙𝟏 = −𝟑−−𝟐 𝟏−𝟑 = −𝟏 −𝟐 = 𝟏 𝟐
We now use the slope and a point to find the y intercept (b).
y < mx + b
-3 = 𝟏 𝟐 (𝟏) + b
-3 - 𝟏 𝟐 = b
Isolate b:
b = - 𝟔 𝟐 − 𝟏 𝟐 = - 𝟕 𝟐

22. y < 𝟏 𝟐 x -3.5 y = mx + b m = 𝟏 𝟐 and b = - 𝟕 𝟐
Going back to the equation:
y = mx + b
we replace what we have found:
m = 𝟏 𝟐 and b = - 𝟕 𝟐
To get the final slope-intercept form of the line passing through (3, -2) and(1, -3)
y < 𝟏 𝟐 x -3.5

23. Y-intercept y crossing
We now proceed to graph the equation:
y < 𝟏 𝟐 x - 𝟕 𝟐
Y-intercept y crossing
𝑹𝒊𝒔𝒆 𝑹𝒖𝒏 2 1

24. Real-World:
A fish market charges $9 per pound for cod and $12 per pound per flounder. Let x = pounds of cod and y = pounds of flounder. What is the inequality that shows how much of each type of fish the store must sell per day to reach a daily quota of at least $120?

25. Real-World(SOLUTION):
A fish market charges $9 per pound for cod and $12 per pound per flounder. Let x = pounds of cod and y = pounds of flounder. What is the inequality that shows how much of each type of fish the store must sell per day to reach a daily quota of at least $120?
Cod  x
Flounder  y
At least $120  9x + 12y ≥ 120

26. Any point in the line or in the shaded region is a solution.
y ≥ - 𝟑 𝟒 x + 10
9x + 12y ≥ 120  10 9 8 10 8 4 4 8 10 12
Any point in the line or in the shaded region is a solution.

27. YOU TRY IT:
A music store sells used CDs for $5 and buys used CDs for $1.50. You go to the store with $20 and some CDs to sell. You want to have at least $10 left when you leave the store. Write and graph an inequality to show how many CDs you could buy and sell.

28. Real-World(SOLUTION):
A music store sells used CDs for $5 and buys used CDs for $1.50. You go to the store with $20 and some CDs to sell. You want to have at least $10 left when you leave the store. Write and graph an inequality to show how many CDs you could buy and sell.
Bought CDs  -5x
Sold CDs  +1.5y
At least $10 left  -5x + 1.5y ≥ -10
NOTE: -10 since you spent this money.

29. Any point in the line or in the shaded region is a solution.
y ≥ 𝟓 𝟏.𝟓 x – 6.6
-5x + 1.5y ≥ -10  10 8 6 4 2 1 2 3 4
Any point in the line or in the shaded region is a solution.

30. VIDEOS: Linear Inequalities

31. CLASSWORK: Page
Problems: As many as needed to master the
concept.