Transcript Solve the linear system by graphing

3.1 Solving Linear Systems by graphing
What you should learn:
Goal 1 Graph and solve systems of linear
equations in two variables.
Goal 2 Use linear systems to solve real-life
problems.
3.1 Solving Linear Systems by graphing
Example 1)
Decide whether the ordered pairs are solution
3x  2 y  4
 x  3 y  5
{
(2, -1), (3, 0)
3.1 Solving Linear Systems by graphing
Example 2)
Solve the linear system by graphing
{
4 x  2 y  12
2 x  5 y  10
1.Graph each line.
2. Use any method to graph the lines.
3.Where they intersect is the solution
3.1 Solving Linear Systems by graphing
Solve the linear system by graphing
{
4 x  2 y  12
2 x  5 y  10
4 x  2 y  12 2 x  5 y  10
x-int = 3
y-int = 6
x-int = 5
y-int = 2
3.1 Solving Linear Systems by graphing
example 3)
Solve the linear system by graphing
{
3
y   x3
4
4x  2 y  8
1.Graph each line.
2. Use any method to graph the lines.
3.Where they intersect is the solution
3.1 Solving Linear Systems by graphing
Graph the line
3
y   x3
4
Slope =  3
4
y- intercept = 3
or
(0,3)
Graph the line
4x  2 y  8
x-int = 2
y-int = -4
Where do they intersect?
3.1 Solving Linear Systems by graphing
3.1 Solving Linear Systems by graphing
Example 1)
Your friends are planning a 7 day trip to Florida. You estimate that it
will cost $275 per day in Tampa and $400 per day in Orlando. Your
total budget for the 7 days is $2300. How many day should you
spend in each location?
1. Set-up 2 linear equations.
-(1) for days
-(1) for money
x y 7
275 x  400 y  2300
Reflection on the Section
Explain how to use a graph to determine how many
solutions there are for a system of linear equations.
assignment
3.1 solving linear systems by graphing
3.2 Solving Linear Systems Algebraically
What you should learn:
Goal 1 Use algebraic method to solve systems
of linear equations .
Substitution Method and
Combination Method
Goal 2 Use linear systems to solve real-life
problems.
3.2 Solving Linear Systems Algebraically
Solve the linear system by substitution
example 1)
{
 x  y 1
2 x  y  2
1. Solve either equation for either one of
its variables.
2. Substitute this expression into the other
equation and solve for the other variable.
3. Substitute this value into the revised
first equation and solve.
4. Check the solution pair in each of the
original equations.
3.2 Solving Linear Systems Algebraically
Solve the linear system by substitution
 x  y 1
2 x  y  2
Begin by solving Equation 1 for y
equation 1
{
 x  y 1
+x
+x
y  x 1
Substitute this expression for y in Equation 2
and solve for x.
3.2 Solving Linear Systems Algebraically
equation 2
2 x  y  2
2 x  ( x  1)  2
3x  1  2
-1
-1
3x  3
x  1
Substitute this value into the revised first equation
and solve for y
3.2 Solving Linear Systems Algebraically
x  1
y  x 1
y  1 1
y0
The solution is (-1, 0). Check to see that it
satisfies each of the original equations.
3.2 Solving Linear Systems Algebraically
Solve the linear system by substitution
example 2)
2x  2 y  3
x  4 y  1
Begin by solving Equation 2 for x
{
x  4 y  1
+ 4y + 4y
x  4 y 1
Substitute this expression for x in Equation 1
and solve for y.
3.2 Solving Linear Systems Algebraically
2x  2 y  3
2(4 y  1)  2 y  3
8y  2  2y  3
10 y  2  3
+2 +2
10 y  5
1
y
2
Substitute this value
into the revised first
equation and solve for x
3.2 Solving Linear Systems Algebraically
1
y
2
x  4 y 1
1
x  4   1
2
x 1
The solution is (1, 1/2). Check to see that it
satisfies each of the original equations.
3.2 Solving Linear Systems Algebraically
Modeling Real – Life situation
You are selling tickets at a high school
football game. Student tickets cost $2 and
general admission tickets cost $3. You sell
1957 tickets and collect $5035. How many
of each type of ticket did you sell?
example 3)
Students
2x
General
3y
Total $ collected 5035
3.2 Solving Linear Systems Algebraically
2 x  3 y  5035
The other equation is the # of student
plus the # of general tickets is 1957.
{
x  y  1957
2 x  3 y  5035
Let’s solve Equation 1 for y
y   x  1957
3.2 Solving Linear Systems Algebraically
Substitute in Equation 2 for y and solve for x
2 x  3( x  1957)  5035
2x  3x  5871  5035
 x  5871 5035
 x  836
x  836 # of student tickets
y  (836)  1957
# of general tickets
y  1121
3.2 Solving Linear Systems Algebraically
Solve the linear system by Linear Combination
1. Arrange the equations with like terms in
columns.
2. Study the coefficients of x (or y).
3. Multiply one or both equations by an
appropriate number to obtain new coefficients
for x (or y) that are opposites.
4.Add the equations and solve for the remaining
variable.
5. substitute the value obtained in Step 4 into
either of the original equations and solve for
the other variable.
3.2 Solving Linear Systems Algebraically
Solve the linear system by Linear Combination
Example 1)
{
4x  3y  1
2x  3y  1
In this linear system, the coefficients for y are
opposites.
By adding the two equations, you obtain and
equation that has only one variable, x.
3.2 Solving Linear Systems Algebraically
Solve the linear system by Linear Combination
Example 1)
{
+
4x  3y  1
2x  3y  1
6x  2
6
6
1
x
3
Now, we take this value and sub it in
for either equation…
3.2 Solving Linear Systems Algebraically
I used the 1st equation…
4x  3y  1
1
4   3 y  1
3
1
3y  
3
1
y
9
So, our solution is
1 1
 , 
3 9
3.2 Solving Linear Systems Algebraically
Solve the linear system by Linear Combination
Example 2)
3x  5 y  6
 4x  2 y  5
{
(4) 3x  5 y  6
(3)  4x  2 y  5
12 x  20 y  24
 12 x  6 y  15
26 y  39
Now, take this
and sub it in…
3
y
2
3.2 Solving Linear Systems Algebraically
 4x  2 y  5
3
 4 x  2   5
2
 4x  3  5
 4x  2
1
x
2
So, our solution is
 1 3
 , 
 2 2
3.2 Solving Linear Systems Algebraically
Solve the linear system with Many Solutions
example 1)
{
2x  y  3
4x  2 y  6
You can use any of the three methods to
show that the system has many solutions.
-2 ( 2 x  y  3 )
4x  2 y  6
 4 x  2 y  6
4x  2 y  6
0=0
This means all numbers work.
3.2 Solving Linear Systems Algebraically
Solve the linear system with Many Solutions
{
2x  y  3
4x  2 y  6
If we graphed
these two lines,
they will be the
same line.
3.2 Solving Linear Systems Algebraically
Solve the linear system with NO Solutions
example 2)
{
2x  y  3
2 x  y  1
You can use any of the three methods to
show that the system has no solutions.
2x  y  3
2 x  y  1
0=4
This means no numbers work.
3.2 Solving Linear Systems Algebraically
Solve the linear system with NO Solutions
{
2x  y  3
2 x  y  1
If we graphed
these two lines,
they will be
parallel lines.
3.2 Solving Linear Systems Algebraically
Parallel Lines – the slopes are the same
the y-intercepts are different
Same lines – the slopes are the same
the y-intercepts are the same.
Lines are either Parallel, the same, or they
just intersect somewhere.
Perpendicular Lines –
*slopes are opposite inverses
y-intercepts are different.
3.2 Solving Linear Systems Algebraically
Example 1)
In one week a music store sold 7 violins for a total of
$1600. Two different types of violins were sold. One
type cost $200 and the other type cost $300. How
many of each type of violin did the store sell?
What to do? Set-up two equations.
equation 1
equation 2
# of
# of
+
type A
type B
Price of
type A
# of
. type
A
=
total # sold
Price of
+
type B
# of
. type
B
=
Total sales
equation 1
equation 2
# of
# of
+
type A
type B
Price of
type A
# of
. type
A
=
total # sold
Price of
+
type B
# of
. type
B
=
Total sales
equation 1
equation 2
# of
# of
+
type A
type B
Price of
type A
# of
. type
A
=
total # sold
Price of
+
type B
# of
. type
B
=
Total sales
x y 7
200 x  300 y  1600
Which method would you like to use to
solve the linear system?
Example 2)
You have $3.55 of change in a bag. There are dimes
and quarters only. If there is a total of 22 coins. How
many of each are there?
Set-up two equations.
equation 1
equation 2
# of
# of =
+
dimes
Quarters
0.10
# of
. dimes
+
total # coins
0.25
# of
. quarters
Total change
=
# of
# of =
+
dimes
Quarters
0.10
.
# of
+
dimes
0.25
.
total # coins
Total
# of
=
quarters
change
D + Q = 22
0.10D + 0.25Q = 3.55
(-0.10)
D + Q = 22
0.10D + 0.25Q = 3.55
-0.10D - 0.10Q = -2.2
0.10D + 0.25Q = 3.55
0.15Q = 1.35
0.15
0.15
Q = 9
Example 3)
A music Club A has a $30 membership fee and
charges $8 per CD. Music club B has a $45
membership fee and charges $7 per CD. How many
CD’s would a person need to buy for Club A to more
expensive than Club B?
What to do? Set-up two equations.
expression A
Membership
+
fee
Per
CD
expression B
Membership +
fee
Per
CD
Set these equal to eachother.
Membership
+
fee
Per
CD
Membership +
fee
Per
CD
Membership
+
fee
Per
CD
=
Membership +
fee
Per
CD
30 + 8x = 45 + 7x
-7x
-7x
30 + x = 45
-30
-30
x = 15
So, they are
equally as good
at 15 CD’s.
How many CD’s would a person need to buy for
Club A to more expensive than Club B? 16 or more
Example 2)
You combine 2 solutions to form a mixture that is 40%
acid.
One solution is 20% acid and the other is 50% acid.
If you have 90 milliliters of the mixture, how much of
each solution was used to create the mixture?
Let’s look at we have…
Try setting-up 2 equations.
One equation that has the amount of 1st solution
plus amount of 2nd solution.
Setting-up equations…let’s say that
x  y  90
x is the amount of 1st solution
y is the amount of 2nd solution
90 is total amount
0.2 x  0.5 y  0.4(90)
Next equation should
be the % of acid times
the amount each
solution.
x  y  90
0.2 x  0.5 y  0.4(90)
Solve using any
method you like.
Probably substitution.
Reflection on the Section
When using the linear combination method for
solving a linear system, why would you want to
have the coefficients of one of the variables be
opposites?
assignment
3.2 solving linear systems by algebraically
3.3 Solving Linear Systems of Linear Inequalities
What you should learn:
Goal 1 Graph a system of linear inequalities
to find the solutions of the system.
Goal 2 Use linear systems to solve real-life
problems.
3.3 Solving Linear Systems of Linear Inequalities
example 1) Graph the system of linear inequalities
{
3x  y  3
y  x 1
What we are going to do is graph and
shade each inequality separately, but
both on the same coordinated plane.
33 Graphing and Solving Systems of Linear Inequalities
Sketch the graph of …
3x  y  3
Graph the line
3x  y  3
either solve for y,
(slope-intercept form)
find x- and y-int
Test an easy to deal with
point…like (0,0)
or
3(0)  (0)  3
03
False
So, shade in the
other side.
3.3 Graphing and Solving Systems of Linear Inequalities
sketch the graph of …
y  x 1
Graph the line
y  x 1
either use the slope and y-intercept,
(slope-intercept form)
or find x- and y-int
Test an easy to deal with
point…like (0,0)
(0)  (0)  1
0 1
False
So, shade in the
other side.
33 Graphing and Solving Systems of Linear Inequalities
When both are graphed and shaded, the solution is
the doubled shaded area.
33 Graphing and Solving Systems of Linear Inequalities
example 2 ) sketch the graph of …
{
y2
x  1
y  x2
3.3 Graphing and Solving Systems of Linear Inequalities
example ) sketch the graph of …
{
y2
3.3 Graphing and Solving Systems of Linear Inequalities
x  1
3.3 Graphing and Solving Systems of Linear Inequalities
y  x2
33 Graphing and Solving Systems of Linear Inequalities
Reflection on the Section
What is the procedure used to graph a system of
linear inequalities?
assignment
3.3 solving linear systems of Inequalities
3.4 Linear Programming
What you should learn:
Goal 1 Solve linear programming problems.
Goal 2 Use linear systems to solve real-life
problems.
3.4 Linear Programming
Reflection on the Section
Why do you need to find the vertices of the
feasible region when using linear programming?
assignment
3.4 linear Programming
3.5 Graphing Linear Equations in Three Variables.
What you should learn:
Goal 1 Graph linear equations in three
variables and evaluate linear functions
of two variables.
Goal 2 Use functions of two variables to
solve real-life problems.
3.5 Graphing Linear Equations in Three Variables
Graphing in Three Dimensions
Solutions of equations in three variables can be pictured with a
three-dimensional coordinate system.
To construct such a system, begin with the xy-coordinate plane in a horizontal
position. Then draw the z-axis as a vertical line through the origin.
In much the same way that points in a two
dimensional coordinate system are represented by
ordered pairs, each point in space can be represented
by an ordered triple (x, y, z).
Drawing the point represented by an ordered triple is
called plotting the point.
The three axes, taken two at a time, determine three coordinate planes that
divide space into eight octants.
The first octant is one for which all three coordinates are positive.
Plotting Points in Three Dimensions
Plot the ordered triple in a three-dimensional coordinate system.
(–5, 3, 4)
SOLUTION
To plot (–5, 3, 4), it helps to first find
the point (–5, 3) in the xy-plane.
The point (–5, 3, 4), lies four units
above.
Plotting Points in Three Dimensions
Plot the ordered triple in a three-dimensional coordinate system.
(–5, 3, 4)
(3, – 4, –2)
SOLUTION
SOLUTION
To plot (–5, 3, 4), it helps to first find
the point (–5, 3) in the xy-plane.
The point (–5, 3, 4), lies four units
above.
To plot (3, – 4, –2) it helps to first find
the point (3, – 4) in the xy-plane.
The point (3, – 4, –2) lies two units
below.
Graphing in Three Dimensions
A linear equation in three variables x, y, z is an equation in the form
ax + by + cz = d
where a, b, and c are not all 0.
An ordered triple (x, y, z) is a solution of this equation if the equation is true
when the values of x, y, and z are substituted into the equation.
The graph of an equation in three variables is the graph of all its solutions.
The graph of a linear equation in three variables is a plane.
A linear equation in x, y, and z can be written as a function of two variables.
To do this, solve the equation for z. Then replace z with f(x, y).
Graphing a Linear Equation in Three Variables
Sketch the graph of 3x + 2y + 4z = 12.
SOLUTION
Begin by finding the points at which the graph intersects the axes.
Let x = 0, and y = 0, and solve for z to get z = 3.
This tells you that the z-intercept is 3, so plot the
point (0, 0, 3).
In a similar way, you can find the x-intercept is
4 and the y-intercept is 6.
After plotting (0, 0, 3), (4, 0, 0) and (0, 6, 0), you can connect these points with
lines to form the triangular region of the plane that lies in the first octant.
Evaluating Functions of Two Variables
Write the linear equation 3x + 2y + 4z = 12 as a function of x and y.
Evaluate the function when x = 1 and y = 3. Interpret the result geometrically.
SOLUTION
3x + 2y + 4z = 12
4z = 12 –3x –2y
Write original function.
Isolate z-term.
z=
1
(12 – 3x – 2y)
4
Solve for z.
f(x, y) =
1
(12 – 3x – 2y)
4
Replace z with f(x, y).
f(1, 3) =
1
3
(12 – 3(1) – 2(3)) =
4
4
Evaluate when x = 1 and y = 3.
(
3
This tells you that the graph of f contains the point 1, 3,
4
).
Using Functions of Two Variables in Real Life
Landscaping You are planting a lawn and decide to use a mixture of two
types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound
and the rye costs $1.50 per pound. To spread the seed you buy a spreader
that costs $35.
Write a model for the total amount you will spend as a function of the number
of pounds of bluegrass and rye.
SOLUTION
Your total cost involves two variable costs (for the two types of seed) and one
fixed cost (for the spreader).
Verbal Model
…
Total
Bluegrass
=
•
cost
cost
Bluegrass
+
amount
Rye
•
cost
Rye
Spreader
+
amount
cost
Modeling a Real-Life Situation
Landscaping You are planting a lawn and decide to use a mixture of two
types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound
and the rye costs $1.50 per pound. To spread the seed you buy a spreader
that costs $35.
Write a model for the total amount you will spend as a function of the number
of pounds of bluegrass and rye.
…
Labels
…
Total cost = C
Bluegrass cost = 2
(dollars)
(dollars per pound)
Bluegrass amount = x
Rye cost = 1.5
Rye amount = y
Spreader cost = 35
(pounds)
(dollars per pound)
(pounds)
(dollars)
Modeling a Real-Life Situation
Landscaping You are planting a lawn and decide to use a mixture of two
types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound
and the rye costs $1.50 per pound. To spread the seed you buy a spreader
that costs $35.
Write a model for the total amount you will spend as a function of the number
of pounds of bluegrass and rye.
…
Algebraic
Model
Total
=
cost
Bluegrass
cost
•
Bluegrass
amount
+
Rye
cost
•
Rye
amount
+
Spreader
cost
C = 2 x + 1.5y + 35
Evaluate the model for several different amounts of bluegrass and rye, and
organize your results in a table.
To evaluate the function of two variables, substitute values of x and y into the
function.
Modeling a Real-Life Situation
Landscaping You are planting a lawn and decide to use a mixture of two
types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound
and the rye costs $1.50 per pound. To spread the seed you buy a spreader
that costs $35.
Evaluate the model for several different amounts of bluegrass and rye, and
organize your results in a table.
For instance, when x = 10 and y = 20, then the total cost is:
C = 2 x + 1.5 y + 35
Write original function.
C = 2 (10) + 1.5(20) + 35
Substitute for x and y.
= 85
Simplify.
Modeling a Real-Life Situation
Landscaping You are planting a lawn and decide to use a mixture of two
types of grass seed: bluegrass and rye. The bluegrass costs $2 per pound
and the rye costs $1.50 per pound. To spread the seed you buy a spreader
that costs $35.
Evaluate the model for several different amounts of bluegrass and rye, and
organize your results in a table.
The table shows the total cost for several different values of x and y.
x Bluegrass (lb)
y Rye (lb)
0
10
20
30
40
10
$70
$85
$100
$115
20
$90
$105
$120
$135
30
$110
$125
$140
$155
40
$130
$145
$160
$175
Reflection on the Section
Name a point on the graph of f if f(-2,3) = 7
assignment
3.5 Graphing Linear Equations in Three Variables
3.6 Solving Systems of Linear Equations in Three Variables.
What you should learn:
Goal 1 Solve systems of linear equations in
three variables.
Goal 2 Use linear systems in 3 variables to
solve real-life problems.
3.6 Solving Systems of Linear Equations in 3 variables
System of three linear equations looks like this:
x  2 y  3 z  3
2 x  5 y  4 z  13
5x  4 y  z  5
We can solve by using: Linear Combination Method
Solutions can be one, none, or many.
They are written as an ordered triple (x, y, z)
Solve the system.
Ex)
3x  2 y  4 z  11
2 x  y  3z  4
5 x  3 y  5 z  1
3x  2 y  4 z  11
(2) 2 x  y  3z  4
1 Eliminate one of the variables
in two of the original equations.
Add 2 times the second
equation to the first.
Solve the system.
Ex)
3x  2 y  4 z  11
2 x  y  3z  4
5 x  3 y  5 z  1
3x  2 y  4 z  11
4x  2 y  6z  8
7 x  10z  19
1 Eliminate one of the variables
in two of the original equations.
Add 2 times the second
equation to the first.
Eliminate the same variable by
adding two other equations.
Solve the system.
Ex)
3x  2 y  4 z  11
2 x  y  3z  4
5 x  3 y  5 z  1
(-3) 2 x  y  3z  4
5 x  3 y  5 z  1
1 Eliminate one of the variables
in two of the original equations.
Add -3 times the first
equation to the second.
Solve the system.
Ex)
3x  2 y  4 z  11
2 x  y  3z  4
5 x  3 y  5 z  1
 6 x  3 y  9 z  12
5 x  3 y  5 z  1
 x  4z  13
1 Eliminate one of the variables
in two of the original equations.
Add -3 times the first
equation to the second.
Now take those 2 equations …
7 x  10z  19
 x  4z  13
2 Solve the new system.
7 x  10z  19
 7 x  28z  91
18z  72
-18
Add 7 times the second
equation to the first.
Solve for z.
-18
z=4
Substitute into new 1 or 2.
z=4
7 x  10(4)  19
x = -3
2(3)  y  3(4)  4
3 Substitute x = -3 and z = 4 into
an original equation and solve
for y
Solve for y.
y=2
The solution is x = -3, y = 2,
and z = 4, or the triple ordered
pair (-3, 2, 4).
Reflection on the Section
How do you decide if a linear system in 3
variables has one, infinitely many, or no solutions
using the equations? Using the graphs?
assignment
3.6 Solving Systems of Linear Equations in 3 variables