Transcript Power Point Version

College Algebra Week 4
Chapter 8.1, 8.2
The University of Phoenix
Inst. John Ensworth
Short section – 8.1 and 8.2
• In these two sections, we work a bit more
with situations where we have two
equations.
• We want to find if and where they meet!
• We want to know what (x,y) they share.
• This is LIKE our overlapping shading, but
we’re just looking for one point!
Section 8.1 Graphing and
Substituting to find that special
(x,y)
• We’ll take two equations, graph them and
see where they cross…
Example 1 A system with only
one solution
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y=x+2 done… m=1 and (0,2)
x+y=4  y=-x+4 m=-1 and (0,4)
Plot it!
It looks like
(1,3)
Let’s check it!
Checking Example 1
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(1,3) Plug into the origional equations
y=x+2
x+y=4
3=1+2 AND 1+3=4
3=3
AND
4=4
Both at the same time and both right!
GREAT!
Example 2 another graphing
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2x-3y=6
and
3y-2x=3
Get that y=mx+b form!
-3y=-2x+6
and
3y=2x+3
y=2/3 x –2
and
x= 2/3 y +1
Giving us graph fodder…
=2/3 and (0,-2)
and m=2/3 and (0,1)
Example 2 the graph
• Oh oh!
• No answer
• Null Set
Example 3 Another by graphing
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2(y+2) = x
and
x-2y=4
Get that y=mx+b form!
2y+4 =x
and
-2y=-x+4
2y=x-4
and
y= ½ x –2
y= ½ x –2
and
y= ½ x –2
Giving us graph fodder…
=1/2 and (0,-2)
and m=1/2 and (0,-2)
Something strange going on?
Example 3 the graph and solution
• They are the
SAME!
• There are an
INFINITE
number of
solutions
• {(x,y)|x-2y=4}
Definitions!
• So we say we have a system of equations that are
independent, inconsistent or dependent based on
how many solutions there are.
A quicker method
• The graphing method we’ve used is clunky.
You have to make a graph every time.
What if the answer is 8.394349 ? Can you
read the graph THAT closely?
• So we’ll try another method… we’ll use the
“Substitution Method”
• (There must be some substituting of some
sort going on here!)
Example 4 Independent system
solved by substitution
• 2x+3y=8
• y=-2x+6
• Notice that the second equation tells us
what y equals. Why not stick it INTO the y
in the first equation?
• Are we crazy enough to do THAT?
Ex 4 Substituting
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2x+3y=8
y=-2x+6
2x+3( -2x+6)=8
2x-6x+18=8
-4x+18=8
-4x=-10
x= 5/2 we have half of the answer, what is y?
Ex 4 finding y
• x= 5/2 we can use our second equation
y= -2x+6
y= -2(5/2) +6 = -5+6 = 1 So we have (5/2,1)
• Check it!
• 2(5/2) + 3(1) = 8
and 1=-2(5/2)+6
• 5+3 = 8
and 1=-6+6 = 1
• Check
and
Check!
Ex 5 How about substituting and
finding an inconsistent solution
(no answer)?
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x-2y=3
2x-4y=7
It’s easiest to solve for x in the first equation
x= 2y+3
Now stick that into the second equation
2(2y+3)-4y=7
4y+6-4y = 7
6=7 BUZZ! It crashed and burned, it’s inconsistent.
Ex 6 A dependent system solved
by substitution
(all numbers = answers)
• 2x+3y=5+x+4y
• y=x-5
• We have y ready to use so plug it into the
first equation
• 2x + 3(x-5) = 5+x+4(x-5)
• 2x+3x-15=5+x+4x-20
• 5x-15=5x-15 Oh, oh, they are the SAME!
All answers work! It’s dependent.
The cookbook…
Ex 7 An application
• Let’s say we have a swimming pool who’s
rectangular length is 2X it’s width. If the
perimeter is 120 feet, what is it’s Length (L)
and Width (W)
Ex 7 continued
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It told us:
L=2W
And 2L+2W=120
Here we go… we have L so stick it into the
second equation
2(2W)+2W=120
4W+2W=120
6W=120
W=20 feet. That means (using the first
equation) L=2(W) or 2(20) or 40 L=40 feet.
Ex 8 Application - Investments
• Belinda had $20,000 to invest. She invested
part at 10% and the rest at 20%. Her
income was $2160. How much did she
invest in both?
• Summary
Amount
Rate
Interest
First
X
investment
10%
0.1x
Second
Y
investment
12%
0.12y
Ex 8 continued
• x+y=20000
• 0.10x+0.12y=2160
• We’ll solve the first equation for x for simplicity
x=20000-y then stick it into the second equation…
• 0.10(20000-y) + 0.12y = 2160
• 2000 –0.1y +0.12y = 2160
• 0.02y = 2160
• y=8000 then plug that into the first
x+8000=20000 means x=12,000
• So she invested $12,000 at 10% and $8000 at 12%
Exercises 8.1
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Here we go again!
Definitions Q 1-6
Solve by graphing Q7-16
Match graphs to system of equations Q17-20
Solve using the substitution method Q21-44
Application problems Q45-58
Section 8.2 the last of Algebra I
• Now we’ll keep solving systems of linear
equations, but using a new trick.
• This one keeps you from having to
substitute big nasty expressions
• We eliminate a variable by adding or
subtracting the two equations just right.
The Addition Method
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Example 1
3x-5y = -9
4x+5y=23
Notice that if you added –5y to 5y, they kill
each other!
The addition method continued
3x-5y = -9
+ 4x+5y= 23
________________
7x + 0 = 14
solve 7x=14
x=2
Getting y now in the addition
method (Ex 1 continued)
• We know x=2 so plug it into EITHER original
equation!
• 3x-5y = -9
4x+5y=23
• 3(2)-5y = -9
4(2) +5y=23
• 6-5y = -9
8 +5y = 23
• -5y = -15
5y=15
• y=3
y=3
• It worked both ways! So the solution is {(2,3)}
Example 2: What if the equations
don’t help us… make them help us!
• 2x-3y = -13
• 5x-12y=-46
• Nothing will vanish if we add or subtract
these two equations.
• So let’s MAKE them cancel somehow.
• Look at the y’s if the first equation were
multiplied on both sides by –4, what would
happen?
Example 2 continued
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2x-3y = -13
5x-12y=-46
So step one is multiply the first one by –4
(-4)2x-(-4)3y=(-4)(-13)
-8x+12y=52
Now well add them
-8x+12y=52
+ 5x –12y=-46
________________
-3x + 0y = 6
x= -2
Example 2 getting y
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We know x=-2
Plug it into either of the first equations again and get X
2x-3y=-13
2(-2) –3y=-13
-4 –3y = -13
-3y=-9
y=3
So (-2,3) is IT!
• If you want, you can plug it into either or both of the
starting equations and check it. We’ll go on though…
Example 3 Multiplying BOTH to
make things nice
• -2x+3y=6
• 3x- 5y = -11
• There are no simple (smaller) numbers we
can multiply ONE equation by to make it
easily add or subtract from the other.
• But what if we multiplied the first by 3 and
the second by –2 ???
Example 3 continued
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Doing two at once…
3(-2x+3y) = 3(6)
2(3x-5y) = 2(-11)
The next step – simplify
-6x+9y=18
6x-10y=-22 Hey! We’re ready!
Example 3 doing the addition
-6x+9y=18
+
6x-10y=-22
________________
0x –y = -4
so y=4
Plug that into one of the original equations…
-2x+3(4)=6  -2x +12=6  -2x =-6  x=3
• So the (3,4)
Example 4 Here is one with
inconsistent equations
-4y = 5x + 7
+ 4y = -5x + 12 It looks EASY!
________________
0 = 0x + 19
0 = 19 BONK!!
Example 5 – We can do the same
thing with fractions and decimals!
• ½ x – 2/3 y = 7
• 2/3x – ¾ y = 11
• What can we do? We want something to cancel.
We want to turn the fractions in front of either
both x’s or both y’s into integers. (We’re using
all the early tools!)
• How about 6 to the first (making 1/x into 3x)
and 12 to the second (making 2/3x into 8x)?
Example 5 continued
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6[ 1/2x – 2/3y] = 6(7)
12[ 2/3x – 3/4y] = 12(11)
Working both
3x- 4y = 42
8x-9y = 132 nice, everything became an
integer
• NOW can we kill something … lets keep
hitting on the x’s
Example 5 continued
• 3x- 4y = 42
• 8x-9y = 132
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Let’s multiply both
equations by the other
equation’s coefficient of x
-8(3x-4y) = -8(42) note I stuck in a – sign for ease
3(8x-9y) = 3(132)
Multiply them out…
-24x+32y=-336
24x – 27y= 396
Example 5 adding
-24x+32y=-336
+ 24x–27y= 396
___________________ y=12
0x +5y = 60
Plug y=12 into one of our original equations to get x
½ x – 2/3(12) = 7  ½ x-8=7  ½ x = 15  
x= 30 Our answer is (30,12)
The Addition Method Cookbook
Example 6 Applications
• At a café the price for 4 fajita dinners and 3
burrito dinners is $48 . The total price for 3
fajita dinners and 2 burrito dinners is $34.
What is the price for each dinner?
• 4x+3y=48
• 3x+2y=34
Example 6 continued
4x+3y=48
3x+2y=34
multiply both by the
other equation’s x
coefficient (to kill the x’s)
-3(4x+3y) = -3(48) make it – for ease
4(3x+2y) = 4(34)
Multiply this out on the next frame…
Example 6 adding!
-12-9y = -144
+ 12x+8y = 136
-----------------------------0x –y = -8
so y=8
Plug that into 4x+3(8) = 48
4x+24=48  4x=24  x=6
So you have (6,8) as the solution!
Fajita dinners cost $6, Burritos cost $8.
Example 7
• Working with mixing oils, we know the
following:
%fat
Canola Oil
Amount of
oil
X
7
Amount of
fat
0.07x
Corn Oil
Y
14
0.14y
Canola and
Corn Oil
280
11
0.11(280)
which is 30.8
gallons
Example 7
• x+y = 280 (total amounts)
• 0.07x+0.14y = 30.8 (amounts of fat)
What if we multiply the FIRST equation by
-0.07 we get
-0.07x -0.07y = -19.6
0.07x+0.14y = 30.8
Example 7 – to the end
-0.07x -0.07y = -19.6
+
0.07x+0.14y = 30.8
_________________________
0x + 0.07 y = 11.2
So y=11.2/0.07 = 160
Using the first equation we put y =160 in to get x
x+160=280  x=120
It takes 120 gal. of canola and 160 gal. of corn oil to
make Crisco Canola and Corn Oil
Exercises 8.2
• Definitions Q1 – Q6
• Solve using the addition method Q7-36
• Application Q37-50