Systems
Day 1
Systems of Linear Equations
System of Linear Equations:
The solution of the
system of equations
=
two or more linear equations together
the point of intersection of the two
lines (makes both equations true)
Example:
Solve by Graphing
a.) y  2 x  3
y  x 1
Step 1: Graph both equations
on the same coordinate plane
Step 2: Find point of intersection
(2,1)
Step 3: Check Answer
y  2x  3 y  x 1
1  2 1
1  22  3
11
1 43
11
The solution is (2,1)
Example:
Solve by graphing
b.)
y  x5
y  4 x  0
Step 1: Graph both equations
on the same coordinate plane
Step 2: Find point of intersection
(-1,4)
Step 3: Check Answer
y  x5
4  1  5
44
y  4 x
4  41
44
The solution is (-1,4)
Infinite Solutions & No Solutions
A system of linear equations has NO SOLUTION
when the graphs of the equations are parallel
(same slope & different y-intercept).
A system of linear equations has an INFINITE NUMBER OF
SOLUTIONS when the graphs of the equations are the same
line (same slope & y-intercept).
Example:
Solve by graphing
a.) y  2 x  1
2 x  y  1
 2x
 2x
have to put in
slope-intercept form
y  2 x  1
Equations have same slopes & different y-intercepts;
therefore they are parallel
Answer: No Solution
Example:
Solve by graphing
1
b.) y   x  2
2
2x  4 y  8
 2x
 2x
4 y  2 x  8
4
4
have to put in
slope-intercept form
4
1
y  x2
2
Equations have the same slope & same
y-intercept; therefore, they are the
same line
Answer: Infinite # of Solutions
Day 2
Example:
 Solve by Graphing
y  x4
y  4x 1
a.)
Step 1: Graph both equations
on the same coordinate plane
Step 2: Find point of intersection
(1,5)
Step 3: Check Answer
y  4x 1
y  x4
5  4(1)  1
5  1 4
55
55
The solution is (1,5)
Example:
 Solve by Graphing
b.)
x  2 y  10
2 x  4 y  10
x  2 y  10
x
x
2 y   x  10
2
2 2
1
y   x5
2
have to put in
slope-intercept form
2 x  4 y  10
2x
2x
4 y  2 x  10
4
4 4
1
1
y  x2
2
2
Equations have same
slopes & different
y-intercepts;
therefore they are
parallel
Answer:
No Solution
Example:
 Solve by Graphing
y  3x  4
have to put in
12 x  4 y  16 slope-intercept form
c.)
y  3x  4
y  3x  4
12 x  4 y  16
12x
12x
4 y  12 x  16
4
4
4
y  3x  4
Equations have same
slopes & same
y-intercepts;
therefore they are
the same line
Answer:
Infinite #
Solutions
Day 1
Example:
Solve using Substitution
a.) y  4 x  8
y  x7
Step 1: Get one of the variables by itself on
one side of the equation
Step 2: Plug into the OTHER equation &
solve for variable
x  7  4x  8
 4x
 4x
5x  7  8
7 7
5x  1
5 5
Step 3: Plug answer back into EITHER of the original
equations to get 2nd variable
1
1

Solution:  ,7 
5 5
1
x
5
y  x7
1
7
5
1
y7
5
y
Example:
b.) y  4 x  8
y  2 x  10
Step 1: Get one of the variables by itself on
one side of the equation
Step 2: Plug into the OTHER equation &
solve for variable
Step 3: Plug answer back into EITHER of the original
equations to get 2nd variable
Solution: 9,28
2x  10  4x  8
 2x
 2x
10  2x  8
8
8
18  2x
2 2
9 x
y  2 x  10
y  29  10
y  28
Example:
c.) 3 y  2 x  4
 6 x  y  7
 6 x  y  7
 6x
 6x
y  6x  7
Need to get a
variable by itself
Now, our equations are:
3y  2x  4
y  6x  7
36 x  7  2 x  4
18x  21  2x  4
20x  21  4
 21  21
Solution: 1.25,.5
20x  25
20 20
5 or
1.25
x
4
y  6x  7
y  61.25  7
y  7.5  7
y  .5
Example:
d.) 2 x  4 y  6
x  3y  7
x  3y  7
 3y  3y
x  3y  7
Need to get a variable by itself
(doesn’t always have to be y)
Now, our equations are:
2 x  4 y  6
x  3y  7
23 y  7  4 y  6
6 y  14  4 y  6
10 y  14  6
 14
Solution:
1,2
 14
10 y  20
10
10
y  2
x  3y  7
x  3 2  7
x  6  7
x 1
Homework
Worksheet: Solve by Substitution
#1-8
Day 2
Example:
Solve using substitution
a.) 4 x  2 y  8
y  2 x  4
4 x  2 2 x  4  8
4x  4x  8  8
88
Variables cancelled out.
Left with a true statement?
88
TRUE!
Answer: Infinite # of Solutions
b.) 6 x  3 y  6
y  2x  5
6x  32x  5  6
6x  6x 15  6
15  6
Variables cancelled out.
Left with a true statement?
15  6
NOT TRUE!
Answer: NO Solution
Homework
Worksheet: Solve by Substitution
#9-21
Example:
• Your school committee is planning an field trip for 193
students.
• There are eight drivers available and two types of
vehicles, school buses and minivans.
• The school buses seat 51 people each and the minivans
seat 8 people each.
• How many buses and minivans will be needed?
Example:
• You have 11 bills in your wallet, some are $5 bills and
some are $10 bills.
• You have a total of $95 in your wallet.
• How many $5 bills and how many $10 bills do you
have?
Day 1
Example:
Solve using Elimination
a.) 5 x  6 y  32
3 x  6 y  48
3 x  6 y  48
32  6 y  48
6  6 y  48
8x  16
8
8
x2
6
Solution: 2,7 
6
6 y  42
6
6
y7
Step 1: Get one pair of
variables that will cancel
-6y and 6y will cancel
Step 2: Add equations &
solve for remaining variable
Step 3: Plug 1st variable into
EITHER equation to get
2nd variable
Example:
b.) 2 x  5 y  22
10 x  3 y  22
 10 x  25 y  110
10 x  3 y  22
 22 y  132
 22  22
Step 1: Multiply one equation by a # to get a pair of
variables that will cancel
 52x  5 y  22
 10 x  25 y  110
Step 2: Add equations & solve for remaining variable
Step 3: Plug 1st variable into EITHER equation to
get 2nd variable
y  6
Solution: 4,6
10 x  3 y  22
10x  3 6  22
10x 18  22
 18  18
10x  40
10 10
x4
Example:
c.) 4 x  2 y  14
7 x  3 y  8
Have to multiply BOTH equations to
get a variable to cancel
34 x  2 y  14
27 x  3 y  8
12 x  6 y  42
14 x  6 y  16
26x  26
26 26
x 1
12 x  6 y  42
14 x  6 y  16
4 x  2 y  14
41  2 y  14
Can plug back into
ANY of the
equations
Solution: 1,5
4  2 y  14
4
4
2 y  10
2 2
y5
Homework:
Worksheet: Solve by Elimination
#1-8
Day 2
Example:
Solving using elimination
a.) 3 x  6 y  12
2x  4 y  8
b.) x  y  4
3 x  3 y  12
Homework
Worksheet: Solving by Elimination
#9-21
Solving by Substitution
Systems with Three Variables
The graph of any equation in the form
Ax + By + Cz = D is a plane.
The solution of a three-variable system is the
intersection of the three planes.
When the solution of a system of equations in 3 variables
is represented by one point,
you can write it as an ordered triple:
(x, y, z)
(alphabetical order)
Example:
Solve using substitution
 x  2 y  z  4

a.) 4 x  y  2 z  1
2 x  2 y  z  10

Step 1: Choose one equation to solve for one of its
variables
Step 2: Substitute the expression into each of the
other equations.
Step 3: Write the two new equations as a system.
Solve for both variables.
Step 4: Plug those variables to one of the original
equations to get remaining variable.
Example:
Solve using substitution
x  3y  z  6

b.) 2 x  5 y  z  2
 x  y  2 z  7

Step 1: Choose one equation to solve for one of its
variables
Step 2: Substitute the expression into each of the
other equations.
Step 3: Write the two new equations as a system.
Solve for both variables.
Step 4: Plug those variables to one of the original
equations to get remaining variable.
Solving by Elimination
Example:
Solve using elimination
 x  3 y  3z  4

a.) 2 x  3 y  z  15
4 x  3 y  z  19

Step 1: Pair the equations to eliminate y, since the yterms are already additive inverses. Add the
equations.
Step 2: Write the two new equations as a system,
solve for the other two variables.
Step 3: Substitute values in one of the original
equations to solve for last remaining variable.
Example:
Solve using elimination
 x  4 y  5 z  7

b.) 3x  2 y  3z  7
2 x  y  5 z  8

Step 1: Find the LCM for the coefficients of the
variable you want to cancel & multiply the equations.
Step 2: Pair the equations to eliminate y, since the yterms are already additive inverses. Add the
equations.
Step 3: Write the two new equations as a system,
solve for the other two variables.
Step 4: Substitute values in one of the original
equations to solve for last remaining variable.
System of Linear Inequalities
x  3 y  2
Shading: Plug in (0,0)
03
0  2
NOT True!
NOT True!
Shade on side
without (0,0)
Shade on side
without (0,0)
The solution is where
the shading overlaps
Solution
Example:
Solve by Graphing
use slope-int.
y

2
x

5
a.)
form
3 x  4 y  12
x=4
y=3
use x & y
intercepts
Shading: Plug in (0,0)
y  2x  5
0  20  5
0  5
3 x  4 y  12
30  40  12
0  12
True!
True!
Example:
b.) y   x  2
2x  4 y  4
x=2
y=1
Shading: Plug in (0,0)
y  x  2
0  02
02
NOT True!
2x  4 y  4
20  40  4
04
True!
Homework
Worksheet:
Solving Systems of Inequalities w/ 2 Equations
Word Problems
Example:
A zoo keeper wants to fence a rectangular habitat for goats. The length
should be at least 80ft & the distance around it should be no more than
310 ft. What are possible dimensions?
x = width of habitat
y = length of habitat
2 x  2 y  310
x=155
y=155
Shading: Plug in (0,0)
0  80
NOT True!
0  310
Length
y  80
140
100
60
True!
Possible dimensions:
20,100 60,80
20
Means: Width is 20 & Length is 100
Width
20
60
100
140
Example:
Suppose you want to fence in a rectangular garden. The length needs to
be at least 50 ft & the perimeter to be no more than 140 ft. Solve by
graphing.
x = width of garden
y = length of garden
70
2 x  2 y  140
x=70
y=70
Shading: Plug in (0,0)
0  50
0  140
NOT True!
True!
Length
y  50
50
30
10
Width
10
30
50
70
Example:
Solve by Graphing
a.)  y  5 x  6

 y  2x 1
 x  1

Example:
Solve by Graphing
1
b.) 
y  2 x 3

x  7
y  2


Recall:
y   xh k
+ V faces up
- V faces down
+h moves LEFT
-h moves RIGHT
+k moves UP
-k moves DOWN
Example:
Solve by Graphing
c.)  y  x  3

 y  x  2  1
Example:
Solve by Graphing
d.) 
 y  4

 y   x  3  1
Day 1
• Linear Programming: a technique that identifies the
minimum or maximum value of some quantity.
• This quantity is modeled with an objective function.
• Limits on the variables in the objective function are
constraints, written as linear inequalities.
Example:
Suppose you want to buy some tapes & CDs. You can afford as
many as 10 tapes and 7 CDs. You want at least 4 CDs & at least
10 hours of recorded music. Each tape holds about 45 minutes of
music and each CD holds about an hour.
a.) Write a system of inequalities.
as many as 10 tapes
as many as 7 Cds
x = #tapes purchased
y = #CDs purchased
x  10
at least 4 CDs
y7
at least 10 hours
These inequalities model the constraints on x & y.
y4
3
x  y  10
4
b.) Graph the system of inequalities
 x  10
y  7

y  4

 3 x  y  10
 4
3
y   x  10
4
The shaded region in the
graph is the feasible region
& it contains all the points
that satisfy all the
constraints.
Say you buy tapes at $8 each
& CDs at $12 each.
The objective function
for total cost C is
C  8 x  12 y
If your total cost is $140, the
equation would be 140 = 8x +12y,
shown by the yellow line
If your total cost is $112, the
equation would be 112 = 8x +12y,
shown by the purple line
As you can see, graphs of the
objective function for various
values of C are parallel.
Lines closer to the origin (0, 0)
represent lower costs.
The graph closest to the origin that
intersects the feasible region
intersects it at the vertex (8, 4).
The graph farthest from the origin that
intersects the feasible region
intersects it at the vertex (10, 7).
Graphs of an objective function
that represent a maximum or
minimum value intersect a
feasible region at a vertex.
Vertex Principle of Linear
Programming
If there is a maximum or a minimum
value of the linear objective function,
it occurs at one or more vertices of the
feasible region.
Example:
• Find the values of x & y that maximize and minimize
P for the objective function P  3 x  2 y
• What is the value of P at each vertex?
Constraints
3

y

x3

2

 y  x  7
x  0

 y  0
Step 1: Graph the Constraints
Example:
• Find the values of x & y that maximize and minimize
P for the objective function P  3 x  2 y
• What is the value of P at each vertex?
Step 2: Find coordinates for each
vertex
Vertex
A(0, 0)
B (2, 0)
C (4,3)
D (0, 7)
Example:
• Find the values of x & y that maximize and minimize
P for the objective function P  3 x  2 y
• What is the value of P at each vertex?
Step 3: Evaluate P at each vertex
Vertex
A(0, 0)
B (2, 0)
C (4, 3)
D (0, 7)
P  3x  2 y
P  3(0)  2(0)  0
P  3(2)  2(0)  6
P  3(4)  2(3)  18
P  3(0)  2(7)  14
When x = 4 and y = 3,
P has its maximum value of 18.
When x = 0 and y = 0,
P has its minimum value of 0.
Homework:
• Worksheet: Textbook page 138 #1-3
• Graph is already done for you,
just have to do steps #2 & 3
(all work on separate paper)
Word Problem
Example:
• Suppose you are selling cases of mixed nuts and roasted peanuts.
• You can order no more than 500 cans and packages & spend no
more than $600.
• How can you maximize your profit? How much is the maximum
profit?
Continued….
• Define variables
x = # of cases of mix nuts ordered
y = # of cases of roasted peanuts ordered
• Write Constraints
spend no more than $600
12 x  20 y  500
24 x  15 y  600
can we have negatives?
x  0, y  0
no more than a total of 500 cans/packages
Continued…
• Write objective function
• We need to write an equation for the profit, since that is what
we are trying to maximize.
P  18 x  15 y
Continued…
• Graph constraints
40
12 x  20 y  500

24 x  15 y  600
 x  0, y  0

30
20
reduce!
3x  5 y  125

8 x  5 y  200
 x  0, y  0

x  41.7, y  25 10
x  25, y  40
10
20
30
40
Continued…
• Find & test vertices
P  18 x  15 y
A  0, 0  P  18(0)  15(0)  0
40
30
B  25,0 P  18(25)  15(0)  450
C 15,16  P  18(15)  15(16)  510
20
D  0, 25 P  18(0)  15(25)  375 10
You can maximize the profit by selling
15 cases of mixed nuts & 16 cases of
roasted peanuts. The maximum profit
is $510.
10
20
30
40
Homework
• Worksheet 3-4
• Use graph paper!