Chapter 7 Graphing Linear Equations 7.1 -The Cartesian Coordinate System -Linear Equations in two variables -Plot points in the Cartesian Coord.System -Determine whether an ordered pair is a solution of an equation.

7.1 Cartesian Coordinate System This is the Cartesian (also referred to as the Rectangular) Coordinate System Y-axis is the vertical axis X-axis is the horizontal axis Origin-the point where the two axes intersect; has coordinates of (0,0)

Notice that the quadrants are numbered in a counter-clockwise direction with Roman Numerals. All the points in a particular quadrant have the same sign orientation. The right hand picture below shows the signs of all the points in each quadrant. Ordered pairs correspond to points in each quadrant as well as on each axis. (3,5) is an ordered pair. 3 is the x-coordinate and 5 is the y-coordinate. (x,y)

To plot points on the Cartesian Coordinate System: Start at the origin Move right or left according to the x-coordinate rt=(+) and left=(-); Keep your pencil here. 3) From this new location, move up or down according to the y-coordinate; up= (+) and down = (-) 4) Place a dot and label when necessary.

7.1 Linear Equations in two Variables A linear equation in two variables is an equation that can be written in the form ax+by=c where a,b, and c are real numbers. 4x + 3y = 12 x = 5 - 3y 2x = 5y y=7 When we graph equations of this form, we get LINES.

The previous examples are all linear equations, but only one of them is in STANDARD FORM: ax + by =c 4x + 3y = 12 x = 5 - 3y 2x = 5y y=7 Because these equations have two variables, they will have two solutions. – an x and a y given as an ordered pair. (x,y) A solution is something that makes the equation true. One way to find solutions is to take a possible solution and plug it in to see if it makes a true statement.

Consider the point (1,2) as (x,y) 4(1) + 3(2) = 12 4 + 6 = 10 (1,2) is not a solution of this equation. Consider the point (3,0) 4(3) + 3(0) = 12 12 + 0 = 12 (3,0) is a solution of this equation.

While equations with one variable (like we solved in chapter 2) have only one solution, equations in two variables have an infinite number of solutions. Can you find another solution for the equation 4x + 3y = 12? Consider the point (0,4) 4x + 3y = 12 4(0) + 3(4) = 12 0 + 12 = 12 (0,4) is a solution for the equation.

We can find more solutions by: 1)Choosing an x-value and 2) Solving for y (or vice versa) 4x + 3y = 12 Now choose an x-value (1) 4(1) + 3y = 12 by substituting 1 for x 4 + 3y = 12 Now solve for y 3y = 8 y = 8/3 or 2 So if x=1 then y= 2 so the ordered pair (1,2 ) is a solution of the equation.

When points fall in a line they are said to be collinear. If we want to graph an equation, we need at least two ordered pairs, but preferably three. This will show us where our line will be. The line represents all the points that are solutions for this particular equation. (0,4), (3,0), (1,2 ) When points fall in a line they are said to be collinear. These points are collinear so we could draw the line.

7.2 Graphing Linear Equations Choose a value for one of the variables Substitute that value into the equation Solve for the other variable This gives you an ordered pair Repeat the above steps until you have 3 points. Plot the points and see if they are collinear. Draw the line through the points.

3x + 2y = 6 Find point #1 Choose x = 0 3(0) + 2y = 6 2y = 6 y = 3 Choose a value for one of the variables Substitute that value into the equation Solve for the other variable This gives you an ordered pair Repeat the above steps until you have 3 points. Plot the points and see if they are collinear. Draw the line through the points. 3x + 2y = 6 Find point #1 Choose x = 0 3(0) + 2y = 6 2y = 6 y = 3 (0,3)

3x + 2y = 6 Find point #2 Choose x = 1 3(1) + 2y = 6 3 + 2y = 6 2y = 3 Choose a value for one of the variables Substitute that value into the equation Solve for the other variable This gives you an ordered pair Repeat the above steps until you have 3 points. Plot the points and see if they are collinear. Draw the line through the points. 3x + 2y = 6 Find point #2 Choose x = 1 3(1) + 2y = 6 3 + 2y = 6 2y = 3 y = 1 ½ (1, 1 ½ )

3x + 2y = 6 Find point #3 Choose x = -1 3(-1) + 2y = 6 -3 + 2y = 6 (-1, 4 ½ ) Choose a value for one of the variables Substitute that value into the equation Solve for the other variable This gives you an ordered pair Repeat the above steps until you have 3 points. Plot the points and see if they are collinear. Draw the line through the points.

Plot the points and then graph the line 3x + 2y = 6 Plot the points and then graph the line (0,3) (1, 1 ½ ) (-1, 4 ½ )

Graph using intercepts X-intercept is where the line crosses the x-axis. An x-intercept always has the form (x,0) Y-intercept is where the line crosses the y-axis. A y-intercept always has the form (0,y) The book presents this as a totally different method, however, it is just picking a certain x or y value and solving for the other variable. Use this method when you see the problem lends itself to it.

Graph using intercepts 4x + 5y = 20 4(0) + 5y = 20 0 + 5y = 20 5y = 20 Y = 4 4x + 5(0) = 20 4x = 20 x = 5 Choose x=0 and solve for y So the point is (0,4) Choose y = 0 and solve for x So the point is (5,0)

Graph 4x + 5y = 20 Plot the two intercepts (0,4) (5,0) A third point could be graphed as a check point. It is not required

Horizontal and Vertical Lines These are special cases. In these equations, only one variable appears. The equation is saying that no matter what the other variable is, this given variable remains constant. These equations look like x = 3 or y = -5

Horizontal and Vertical Lines Let’s take x = 3 That tells me that x=3 no matter what y is. Y can be equal to anything and x will still be equal to three. X Y 3 -1 3 0 3 5 3 -5

Horizontal and Vertical Lines Let’s take y = -5 That tells me that y=-5 no matter what x is. x can be equal to anything and y will still be equal to -5 X Y -3 -5 1 -5 0 -5 6 -5

7.3 Slope of a Line -represented by the letter m -a number that describes the steepness of a line -read from left to right on a graph (uphill lines are positive; downhill lines are negative) -a ratio of vertical change to horizontal change between any two points on a line -referred to also as rise over run. m change in y y2 – y1 change in x x2 – x1

Find the slope using a graph Count up and over from one point on the line to any 2nd point on the line. Start at (-4,1); Go up three units and right five units and land at (1,4) Vert change 3 horiz change 5 This won’t reduce so the slope is 3/5 5 3

Find the slope using the formula Using the points (-9,1) and (7,4) for the same line And the formula y2 – y1 x2 – x1 4 – 1 3 7- -9 16

Find the slope using the formula y2 – y1 x2 – x1 (5,7) and (1,7) m 7-7 0 0 what kind of line? 5-1 4

Find the slope using the formula y2 – y1 x2 – x1 (1,-2) and (1,3) m 3 - -2 5 undefined 1 – 1 0 What kind of line is this?

Uphill lines have positive slope Downhill lines have negative slope Read the lines from left to right on the graph. + slope - slope

Parallel and Perpendicular lines Parallel lines have equal slopes Perpendicular lines have slopes that are negative reciprocals of each other

7.4 Slope Intercept Form Slope Intercept Form: y = mx + b where m is the slope and b is the y-intercept To put an equation in slope-intercept form, solve for y 2x + 3y = 6 3y = -2x + 6 y = x + 2 Now you can pull out the m and b values m= -2/3 and b = 2

Graph using y=mx+b Now using this info: m= -2/3 and b = 2 We can quickly and easily graph this line y = x + 2 b= 2 means the line cross the y-axis at 2 m= -2/3 means it goes down 2 and right 3 Down 2 Right 3

y = (½) x - 3 To graph a line using the m and b 1) Place a dot at the b-value on y-axis (b=-3) 2) From this point move according to the slope and place a 2nd dot (up 1 and right 2) 3) Draw line

Slope Intercept Form We can also use this form and its resulting information to answer questions about the line – such as is it parallel to another line? Or perpendicular to another line?

If we know certain things about a line, we can write the equation of the line in slope-intercept form. We need to know the slope and the intercept values. Sometimes they will give you slope. If they don’t, do they give you two points so that you can find the slope using the formula? If they do not give us an intercept value, we cannot approximate it using the graph. They must tell us what the b-value is.

7.4 Point Slope Form The point slope form is not as user-friendly as the slope intercept form, but it can be used in instances where y=mx + b cannot be used. The point slope form requires only a point and the slope The point slope form: y - y1 = m(x – x1) where m is the slope and (x1, y1) is any point on the line.

Given (-1,3) and (-2,7), write the equation of the line passing through the two points. Because they do not give us the slope, we must use the formula to find it: m 7 – 3 4 -4 -2 - -1 -1 Now choose either point to be (x1, y1) y - y1 = m(x – x1) y – 7 = -4(x - -2) y – 7 = -4(x + 2) y – 7 = -4x -8 y = -4x -1 in slope-int or 4x + y = -1 in standard

“Easiest Case” #1 m is known b is known Go straight to Y=mx + b Substitute in m and b and you are done! m = -1/2 b = 3 y = mx + b y = (-1/2) x + 3

“Easyish Case” #2 m is unknown b is known but maybe hidden Find the slope using the formula Go straight to Y=mx + b Substitute in m and b and you are done! through (10,3) and (0,-2) m = 3- -2 = 5 = 1 10-0 10 2 y = mx + b y = (1/2) x + -2

“Harder Case” #3 m is known b is unknown a point is given use this form: y – y1 = m (x – x1) rearrange into this form: y = mx + b m = -3 through (2,5) y – y1 = m (x – x1) y – 5 = -3 (x – 2) y – 5 = -3x + 6 y = -3x + 11

“Hardest Case” #4 m and b are unknown two points are given use this form: y – y1 = m (x – x1) then rearrange to this: y = mx + b through (7,4) and (6,3) m = 4-3 = 1 7-6 1 y – y1 = m (x – x1) y – 4 = 1(x – 7) y – 4 = x – 7 y = x - 3

7.5 Graph Linear Inequalities 1) 3x + 2y > 5 3x + 2y = 5 2) 2y = -3x + 5 y = x + 1) Replace > with = for now 2) Graph the line (I like using y=mx+b method) 3) Choose a point clearly on either side of the line; (0,0) is a good one if the line doesn’t pass through this point

>,< are dotted lines >,< are solid lines 4) 3x + 2y > 5 3(0) + 2(0) > 5 5) 0 > 5 is False Shade the opposite side from (0,0) 4)Subsitute the check point into the ORIGINAL inequality to see if it is T or F 5) If it is T, shade the same side of the line. If it is F, shade the opposite side of the line. 6) Check your line >,< are dotted lines >,< are solid lines