# Class Notes - PDF 12 by homers

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```									Class Notes
Graphing Linear equations

Approaches

~      Any two points establish a distinct line. All points on that line are solutions to
the linear (line) equation which describes that line

Cover Method               ‘x’ and ‘y’ intercepts are found.

T   Table                  equation written in ‘y’ form ( y = ___ ) then ‘x’ values are
given to find corresponding values for ‘y’.

Cover method                 Find 'x' and 'y'      intercepts                     3y - 6x = 12

~     Make 'x' equal to 0        (cover the 'x' term)

3y - 6( 0 ) = 12
3y = 12                    y = 4            (0, 4)   is the 'y' intercept

~      Make 'y' equal to 0        (cover the 'y' term)

3( 0 ) - 6x = 12
- 6x = 12                   x = -2          (-2, 0)    is the 'x' intercept

~      Plot and connect the points
3y - 6x = 12
Class Notes
Graphing Linear equations

T     Table           Finding (x, y) points which lie on the line                   3y - 6x = 12

~       Solve the equation for 'y'         ( 'y' form or slope-intercept form)

Steps to writing equation in ‘y’ (slope-intercept form)

3y - 6x = 12

3y - 6x + 6x = 6x + 12                              (add     6x to both sides)

3y = 6x + 12                            (simplify)

(1/3) • 3y = (6x + 12) (1/3)                      (multiply both sides by reciprocal of 3 ... 1/3)

y = 2x + 4                              (simplify using distributive property)

~       Construct        T   Table        (assign values for 'x' to find corresponding values for 'y')

x           y                       point

0           2                       ( 0, 4 )
1            6                       ( 1, 6 )
-1           2                         (-1, 2 )
2            8                       ( 2, 8 )

~      Plot and connect the points                   3y - 6x = 12      or       y = 2x + 4
Class Notes
Graphing Linear equations

Does a point lie on the line ?

~       Remember that only two points are necessary to graph a line and that
all points on that line are solutions to the linear equation which describes that line.

~       To determine whether any point (x, y) lies on the line merely substitute
the ‘x’ and ‘y’ values in the equation.

Example: In th linear (line) equation 3y - 6x = 12

~      Does point (1, 5) lie on the line 3y - 6x = 12 ?

Does 3 (5) - 6 (1) = 12 ?          Does    9 = 12 ?         no

Point (1, 5) does NOT lie on the line 3y - 6x = 12

~       Does point (6, 1) lie on the line 3y - 6x = 12 ?

Does 3 (6) - 6 (1) = 12 ?            Does 12 = 12 ?            yes

Point (1, 5) does lie on the line 3y - 6x = 12

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