1. Objectives Linear Equations and Inequalities
In this lesson you'll be reviewing linear equations and inequalities. How many different forms of a line are there? Can you list them?
You'll be using these equations periodically throughout the rest of this course and they appear again when you study calculus, so they are essential concepts to know well. 
Objectives
1. Write the equation of a line in three forms.
2. Identify key components of a line from a given equation.
3. Express the solution to a linear inequality graphically.

2. Key Terms: Linear Equations and Inequalities
1. linear equation: An equation in which the variable is of degree one.
2. linear inequality: An inequality in which the variable is of degree one.
point-slope form: A form of a linear equation that includes the coordinates
of a point on the line and the line's slope.
slope-intercept form: A form of a linear equation that includes the slope of
the line and the value of the y-intercept.
standard form of a linear equation: A linear equation written in the form
Ax + By = C.
6. Undefined: A value that cannot be computed.

3. Point-Slope Form
The first form of a line we'll look at is perhaps the least popular among students. It is called the point-slope form of a line and, surprisingly, it is probably the easiest form to work with. If that's so, why is it the least popular?

5. Point-Slope Form: y – y1 = m(x – x1)
Write an equation of the line in point-slope form that passes through the given point and has the given slope: (3, 4) , m = ½.
Point-Slope Form: y – y1 = m(x – x1)
Substitute in (x1 , y1) and m into the formula and simplify if possible.
y – y1 = m(x – x1) (3 , 4) m = ½
y – 4 = ½ (x – 3)
Final Answer is y – 4 = ½ (x – 3).

6. Write an equation of the line in point-slope form that passes through the given point and has the given slope: (-9, -6) , m = -2/3.
Point-Slope Form: y – y1 = m(x – x1)
Substitute in (x1 , y1) and m into the formula and simplify if possible.
y – y1 = m(x – x1) (-9 , -6) m = -2/3
y – (-6) = (-2/3)(x – (-9))
y + 6 = (-2/3)(x + 9)
Final Answer is y + 6 = (-2/3)(x + 9).

7. 3. Write an equation in point-slope form of the line that passes through the given points:
(4, -5) (-2 , -7).
First you must find the slope of the line using the slope formula: m = (y2 – y1)/(x2 – x1).
m = (y2 – y1)/(x2 – x1) = (-7 + 5)/(-2 – 4)
= (-2)/(-6) = 1/3
Now use the slope of the line with either point to write your equation in point-slope form.
y – y1 = m(x – x1) (4 , -5) m = 1/3
y + 5 = (1/3)(x – 4) or
y – y1 = m(x – x1) (-2 , -7) m = 1/3
y + 7 = (1/3)(x + 2)
Final Answer is y + 5 = (1/3)(x – 4) or y + 7 = (1/3)(x + 2).

8. Final Answer is y + 7 = (-1)(x – 1) or y + 5 = (-1)(x + 1).
4. Write an equation in point-slope form of the line that passes through the given points:
(1, -7) (-1, -5).
First you must find the slope of the line using the slope formula: m = (y2 – y1)/(x2 – x1).
m = (y2 – y1)/(x2 – x1) = (-5 + 7)/(-1 – 1)
= (2)/(-2) = -1
Now use the slope of the line with either point to write your equation in point-slope form.
y – y1 = m(x – x1) (1 , -7) m = -1
y + 7 = (-1)(x – 1) or
y – y1 = m(x – x1) (-1 , -5) m = 1/3
y + 5 = (-1)(x + 1)
Final Answer is y + 7 = (-1)(x – 1) or y + 5 = (-1)(x + 1).

9. 5. Graph: y – 4 = 2/5(x – 5).
Start by graphing the point (5 , 4) on the graph. Then from that point go up 2 and right 5 to make another point or go down 2 and left 5 to make another point. Then connect the dots with a line.

10. Use either point to write in point-slope form.
6. Write the equation of the line in point-slope form given the two points on the graph: (-2 , 8) and (1 , 2).
Find the slope: m = -3
Use either point to write in point-slope form.
y – 8 = -3(x + 2)
y – 2 = -3(x – 1)

11. Slope-Intercept Form of a Line
It is likely the very first form of a line you learned when you were in algebra: y = mx + b. It makes graphing easy because it provides you with one point, the intercept, and the slope, which helps you plot a second point 1 2

12. To write an equation in slope-intercept form, you will need to find the slope and y-intercept of your line.

13. First locate the y-intercept of the line graphed
First locate the y-intercept of the line graphed. You can see that the line crosses at 2 on the y-axis. Thus b = 2.
Next find the slope of the line. Choose any two points on cross-hairs and count the rise/run to find the slope. I will use the two points (0,2) and (5,4). Thus the slope is m = 2/5.
So the equation of the line in slope-intercept form is: y = (2/5)x + 2.
y = (2/5)x + 2

14. First locate the y-intercept of the line graphed
First locate the y-intercept of the line graphed. You can see that the line crosses at -1 on the y-axis. Thus b = -1.
Next find the slope of the line. Choose any two points on cross-hairs and count the rise/run to find the slope. I will use the two points (0,-1) and (-1,2). Thus the slope is m = -3.
So the equation of the line in slope-intercept form is: y = -3x – 1.
y = -3x – 1

15. Making a Graph from an Equation
Your final practice problem with this concept will be a problem in which the equation is given. Your task will be to graph the line from the equation.
To do this, identify the y-intercept and the slope from the equation; then the line can be graphed using those values.
Plot the y-intercept (b) which is the point (0,b) or the number b on the y-axis.
Then from that point use the slope (m) of the equation to find another point on the line and put a point there.
Finally connect the two points with a line that extends throughout the entire coordinate plane.

16. Graph the equation: y = 3x – 2.
1. Plot the y-intercept (-2) which is the point (0,-2) or the number -2 on the y-axis.
2. Then from that point use the slope (3) of the equation to find another point on the line and put a point there.
3. Finally connect the two points with a line that extends throughout the entire coordinate plane.

17. Graph the equation: y = (-2/3)x + 1.
1. Plot the y-intercept (1) which is the point (0,1) or the number 1 on the y-axis.
2. Then from that point use the slope (-2/3) of the equation to find another point on the line and put a point there.
3. Finally connect the two points with a line that extends throughout the entire coordinate plane.

18. Standard Form of a Linear Equation
It is also often called the "general form" of a line. You may see this form when you are given a linear equation to work with, but you will rarely need to put a line into this form (unless asked!).
The standard form of a line is:
                  
Notice the capital letters A, B, and C in this form. These represent the coefficients and constant terms. In particular, the capital B should help you distinguish the coefficient of y and the lowercase b from slope-intercept form.
The letter A is the coefficient of the x term, B is the coefficient of the y term and C is the constant value. It's important to note that, in standard form, A, B, and C are integer values, which means they aren’t fractions.
All by themselves, these coefficient and constant values don't really have any direct significance to the line, but you can use the values to find two points on your line. And two points are enough information by which to define a line.
But how can you graph an equation in standard form, since it includes no slope, intercept, or points?

20. Find the x-intercept and y-intercept of the equation 3x – 2y = 6.
3x – 2y = 6 Cover up the 2y.
Thus the x-intercept = 2 or the coordinate (2 , 0).
3x – 2y = 6 Cover up the 3x.
Thus the y-intercept = -3 or the coordinate (0, -3).
Plot those two points and draw a line through them. Remember to draw the line through the entire graph.

21. Find the x-intercept and y-intercept of the equation 3x + 4y = 5.
3x + 4y = 5 Cover up the 4y.
Thus the x-intercept = 5/3 or the coordinate (5/3 , 0).
3x + 4y = 5 Cover up the 3x.
Thus the y-intercept = 5/4 or the coordinate (0, 5/4).
Plot those two points and draw a line through them. Remember to draw the line through the entire graph.

22. Vertical and Horizontal Lines
Slope m = 0
Slope m = undefined

24. Stop: Graphing Inequalities next.

25. Graphing Linear Inequalities

28. Graph the equation x = 4. It is a vertical line crossing the x-axis at 4. It is a solid line.
Now shade one side of the line. If you test the point (0 , 0), you see it is a true statement so you shade that side of the line. Or the numbers to the left of 4 are all smaller than 4.

29. y > -4.5
Graph the equation y = It is a horizontal line crossing the y-axis at It is a dashed line.
Now shade one side of the line. If you test the point (0 , 0), you see it is a true statement so you shade that side of the line. Or the numbers above -4.5 are all greater than -4.5.

30. Graph the equation y = 2x – 3
Graph the equation y = 2x – 3. Start at the point (0 , -3) and go up 2 and right one. It is a solid line.
Now shade one side of the line. If you test the point (0 , 0), you see it is a false statement so you shade the opposite side of the line.

31. 5x – 3y > 15
Graph the equation 5x – 3y = 15. Start at the point (0 , -5) and the point (3 , 0) the intercepts of the line. It is a dashed line.
Now shade one side of the line. If you test the point (0 , 0), you see it is a false statement so you shade the opposite side of the line.

32. y < (3/2)x + 2 or 3x – 2y > -4
Write the inequality that the graph below
represents?
y < (3/2)x + 2 or
3x – 2y > -4

33. y ≥ 4x – 1 or 4x – y ≤ 1 8. Write the inequality that the graph below
represents?
y ≥ 4x – 1 or
4x – y ≤ 1

34. y < (1/5)x + 1 or x – 5y > -5
9. Write the inequality that the graph below
represents?
y < (1/5)x + 1 or
x – 5y > -5

35. y < (-1/3)x + 4 or x + 3y < 12
10. Write the inequality that the graph below
represents?
y < (-1/3)x + 4 or
x + 3y < 12