Transcript Welcome to the Unit 2 Seminar for College Algebra!

Welcome to the Unit 4 Seminar
for College Algebra!
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Unit 4 Seminar Agenda
• Slope of a Line
• Writing Equations of Lines
• Solving Systems by Graphing
• Solving Systems Algebraically
Slope
•
•
•
•
4 types of slopes
a diagonal line heading uphill (increasing)
as you look at it from left to right
a diagonal line heading downhill
(decreasing) as you look at it from left to
right
a horizontal line
a vertical line
Finding slope given 2 points
Find the slope between (1,3) and (4,5)
y2  y1
m
x2  x1
53
m
4 1
Example:
Find the slope of the line that passes through the
points (-2, 6) and (-1, -2).
y2  y1
m
x2  x1
6  (2)
m
2  (1)
8
m
 8
1
Example
Find the slope of the line that passes through the
points (0, 3) and (7, -2).
y2  y1
m
x2  x1
3  (2)
m
07
5
m
7
Example
Find the slope of the line that passes
through the points (1, 7) and (-3, 7).
y2  y1
m
x2  x1
77
m
1  (3)
0
m
4
Example
find the slope of the line that passes
through the points (6, 4) and (6, 2).
y2  y1
m
x2  x1
42
m
66
2
m
0
Finding slope and y-intercept
given an equation
• First solve for y
• When you have it in the form y = mx + b
then m is your slope and b is your yintercept
y = -3x + 7
m = -3 is your slope
b = 7 is your y-intercept
y = -3x + 7
y
8
6
4
2
x
-8
-6
-4
-2
2
-2
-4
-6
-8
4
6
8
Example
2 x  4 y  1
2 x  4 y  2 x  2 x  1
4 y  2 x  1
4 y 2 x 1


4
4
4
1
1
y   x
2
4
First move your 2x to the other side
simplify
We need to isolate the y so divide both sides by 4
simplify
Example
x  5y  2
x  5y  x  x  2
5 y  x  2
5 y
x
2


5 5 5
1
2
y   x
5
5
First move your -x to the other side
simplify
We need to isolate the y so divide both sides by- 5
simplify
Parallel and Perpendicular Lines
• Parallel lines have the same slope, so if
one line is y = 2x + 7, then the slope of a
line parallel to this one is also 2.
• Perpendicular lines have “negative
reciprocal” slopes, so if one line is
y = 2x + 7, then the slope of a line
perpendicular would be -1/2
Finding equations of lines using
y = mx + b
Example: Write the equation of a line when the
slope is 2/3 and the y-intercept is (0, 5).
From this information, I know that I will use 2/3
for m and 5 for b.
Slope-intercept form: y = mx + b
y = (2/3)x + (5)
y = (2/3)x + 5 I’m leaving the parentheses
around the 2/3 so that there is no confusion as
to where the x goes (in the numerator).
Finding equations of lines using
y – y1 = m(x – x1)
Example: Write the equation of a line whose slope is -2 and passes
through the point (3, 5).
To prepare for using the point-slope equation, let’s label what we
have:
m = -2, x1 = 3, and y1 = 5. Plug them in.
y – y1 = m(x – x1)
y – (5) = (-2)(x – (3))
y – 5 = -2(x – 3) now distribute through the parentheses
y – 5 = -2x + 6
y – 5 + 5 = -2x + 6 + 5 add 5 to both sides
y = -2x + 11 this is in SLOPE-INTERCEPT form
y + 2x = -2x + 11 + 2x add 2x to both sides in order to get xs and y
all on the same side
2x + y = 11 as in standard form, the x is first, the y is second, then
the equals sign followed by the constant (plain number). This is the
form of the choices on your quiz.
Example
• Example: Write the equation of a line whose
slope is 0 and passes through the point (-2, -2).
• y – y1 = m(x – x1)
• y – (-2) = (0)(x – (-2))
• y + 2 = 0(x + 2) change minus a negative to
plus a positive
• y + 2 = 0 0 times anything is 0!
• y + 2 – 2 = 0 – 2 notice that the x just got wiped
out; therefore, get y alone
• y = -2 this is the equation of a horizontal line
Finding equations of the line
given two points
• First find the slope
• Then use the point-slope equation as
in the last problems.
• If you have the graph of a line, just
pick two points and do as stated
earlier.
Example
Example: write the equation of the line
that passes through the points (4, 2)
and (7, 3).
Step 1: find the slope
y2  y1
m
x2  x1
3 2
m
74
1
m
3
Step 2:
Use the slope and one of
the points (your choice)
and plug them into
POINT-SLOPE form of a
line. (I’m just going to go
with the first point.)
y  y1  m( x  x1 )
1
y  2  ( x  4)
3
1
4
y2 x
3
3
1
4
y22  x 2
3
3
1
2
y  x
3
3
Changing to Standard Form
1 2
y  x  move the x to the other side
3 3
1 1 2 1
y x  x  x
3 3 3 3
1
2
 x  y  multiply by 3 to get rid of the fractions
3
3
2
 1 
(3)   x   (3) y  (3)
3
 3 
x  3y  2
Systems of Equations
• Systems of equations are simply more
than one equation involving more than one
variable for which the values associated
with the variables work for all equations in
the system.
3 Possibilities for Systems of
Equations
1. the two lines INTERSECT
they have ONE point in common
there is ONE unique solution to the system, in the form
(x, y)
2. the two lines are PARALLEL to each other
they have NO points in common
there is NO SOLUTION to the system
3. the two lines are the SAME—called COINCIDENTAL
lines
they have ALL points in common
there are INFINITELY MANY SOLUTIONS to the system
Important Terminology
• DEPENDENT: the graph of two lines looks like
you only graphed one (coincidental lines).
• INDEPENDENT: the graph of two lines shows
up as two lines (parallel and intersecting lines
both qualify).
• CONSISTENT: there is at least one solution to
the system (intersecting lines have one and
coincidental lines have infinitely many).
• INCONSISTENT: there is no solution to the
system (parallel lines).
Solving Systems of Equations
using the Substitution Method
• Step 1: rearrange one equation in terms of
one variable. In other words, get x by
itself or get y by itself, whichever is easier
at the time.
• Step 2: take this result and substitute it
into the OTHER equation.
• Step 3: take this solution, substitute it into
the equation you used in step #1, and
solve for the other variable.
Example
Use the substitution method to find the solution
to the system
x+y=6
3x + 4y = 9
Step 1:
Rearrange one equation in terms of one variable. In
other words, get x by itself or get y by itself,
whichever is easier at the time.
The first equation looks good to rearrange, as the
coefficients of both x and y are 1.
x+y=6
x+y–y=6–y
x=6–y
Step 2:
Take this result and substitute it into the
OTHER equation. In this case, we’re going to
replace x in the other equation with 6 – y.
3x + 4y = 9
3(6 – y) + 4y = 9
Now, it may seem that we just make this
equation longer, but in reality we’re in good
shape. We changed the equation such that
there is now only ONE variable to solve for,
something we know how to do!
18 – 3y + 4y = 9
18 + y = 9
18 + y – 18 = 9 – 18
y = -9
Step 3:
Take this solution, substitute it into the equation you
used in step #1, and solve for the other variable. In this
case, replace y in the step #1 equation with –9.
x+y=6
x + (-9) = 6
x–9=6
x–9+9=6+9
x = 15
Solving Systems of
Equations using the
Elimination/Addition Method
• Step 1: find (because they already exist) or
create (by multiplying) additive inverses of one
variable
• Step 2: combine like terms, straight down the
columns. If you are in standard form as you
should be, the xs are lined up, the ys are lined
up, and the numbers are lined up.
• Step 3: take this solution, substitute it into one of
the original equations, and solve for the other
variable. For this method, it does not matter
which equation you substitute into.
Example
Use the elimination method to find the solution to the
system
x+y=4
x–y=1
**Note: the equations must start in standard form. If
they’re not, rearrange them.
Step 1:
Find (because they already exist) or
create (by multiplying) additive inverses
of one variable. In this case, they
already exist, the ys. (Remember
additive inverses are the same numbers
with different signs like 2 and -2)
Step 2:
Combine like terms, straight down the columns. If you are
in standard form as you should be, the xs are lined up, the
ys are lined up, and the numbers are lined up.
x+y=4
x–y=1
2x = 5
The ys “cancelled out”, leaving only x to solve for. Solve for
x.
2x = 5
2x/2 = 5/2
x = 5/2
Step 3:
Take this solution, substitute it into one of the original
equations, and solve for the other variable. For this
method, it does not matter which equation you
substitute into.
x+y=4
(5/2) + y = 4
5/2 + y – 5/2 = 4 – 5/2
y = 8/2 – 5/2
y = 3/2
Solution: (5/2, 3/2)
Special Cases
1. the two lines INTERSECT
they have ONE point in common
there is ONE unique solution to the system, in the form
(x, y)
2. the two lines are PARALLEL to each other
they have NO points in common
there is NO SOLUTION to the system
3. the two lines are the SAME—called COINCIDENTAL
lines
they have ALL points in common
there are INFINITELY MANY SOLUTIONS to the system
Example: use the elimination method to find the solution to the
system
2x + 5y = 12
2x + 5y = 8
Even though these terms look alike, they are not additive
inverses. Additive inverses have opposite signs.
Let’s say that I choose to eliminate the xs. I need one positive 2x
and one negative 2x. All I have to do to make a –2x is to multiply
one of the equations by –1. I’ll do this to the second equation.
-1[2x + 5y = 8] becomes –2x – 5y = -8
Here’s the system after that step:
2x + 5y = 12
–2x – 5y = -8
Combine like terms:
2x + 5y = 12
–2x – 5y = -8
0=4
Example: use the elimination
method to find the solution to the
system
-3x + 9y = 4
3x – 9y = -4
I’ll make this one easy. We already
have additive inverses of the
xs…and of the ys, as it works out
(hint, hint)…so let’s combine like
terms.
-3x + 9y = 4
3x – 9y = -4
0=0