1 Math 1201 Unit 7: Systems of Linear Equations Read Building On, Big Ideas, and New Vocabulary, p. 392 text. Ch. 7 Notes §7.1 Developing Systems of Linear Equations (1 class) Read Lesson Focus p. 394 text. Outcomes 1. Define a system of linear equations or a linear system. pp. 396, 541 2. Define the solution to a linear system. p. 396 3. Model a situation, using a system of linear equations. pp. 397-400 4. Relate a system of linear equations to the context of a problem. p.400 n Def : A system of linear equations or a linear system is a system of two or more linear equations using the same two variables. E.g.: 0.9 10 15 10 17 11 0.87 ; ; 1 9 2 0.25 0.05 75 8 2 4 a b x y s t x y t s a b Do # 4, p. 401 text in your homework booklet. n Def : The solution to a linear system is the values of the variables that satisfy both equations. E.g.: 6, 4 x y is the solution to 10 2 x y x y because 6 4 10 and 6 4 2 E.g.: Explain why 7, 3 x y is NOT a solution to 10 2 x y x y . For 10, x y . .. 7 3 10 . .. LHS RHS But for 2, x y . .. 7 3 4 . .. LHS RHS Do # 5, p. 401text in your homework booklet.
Transcript
1
Math 1201
Unit 7: Systems of Linear Equations
Read Building On, Big Ideas, and New Vocabulary, p. 392 text.
Ch. 7 Notes
§7.1 Developing Systems of Linear Equations (1 class)
Read Lesson Focus p. 394 text.
Outcomes
1. Define a system of linear equations or a linear system. pp. 396, 541
2. Define the solution to a linear system. p. 396
3. Model a situation, using a system of linear equations. pp. 397-400
4. Relate a system of linear equations to the context of a problem. p.400
nDef : A system of linear equations or a linear system is a system of two or more linear equations
using the same two variables.
E.g.:
0.9 10 1510 17 11 0.87
; ;1 92 0.25 0.05 758
2 4
a bx y s t
x y t sa b
Do # 4, p. 401 text in your homework booklet.
nDef : The solution to a linear system is the values of the variables that satisfy both equations.
E.g.: 6, 4x y is the solution to 10
2
x y
x y
because 6 4 10 and 6 4 2
E.g.: Explain why 7, 3x y is NOT a solution to 10
2
x y
x y
.
For 10,x y
. . . 7 3 10 . . .L H S R H S
But for 2,x y
. . . 7 3 4 . . .L H S R H S
Do # 5, p. 401text in your homework booklet.
2
Using a Diagram to Model a Situation
A volleyball court has a perimeter of 54m. The length of the court is twice the width.
a) Create a linear system to model this situation.
b) Ryan, the V-ball king, states that the court is
18m long and 9m wide. Is he correct?
Let l be the length of the court in metres. Let w be the
width of the court in metres.
a) Since the perimeter of the court is 54m, we can
write the equation:
n2 2 54 Eq 1l w
Since the length is twice the width, we can write the equation:
n2 Eq 2l w
We can combine equations 1 and 2 to make a linear system 2 2 54
2
l w
l w
b) Let 18 and 9l w
. . . 2 18 2 9 36 18 54 . . .L H S R H S
. . . 18 2 9 . . .L H S R H S
This means 18 and 9l w is a solution to 2 2 54
2
l w
l w
. It also means that Ryan, the V-ball king,
is correct.
Do # 8, p 401 text in your homework booklet.
Relating a Linear System to a Problem
E.g.: Create a situation relating to coins that can be modeled by the linear system below and explain the
meaning of each variable.
0.25 0.05 4.50
24
q n
q n
Let q be the number of quarters. Let n be the number of nickels.
Situation: Mikey has a bunch of quarters and nickels in his pocket. He counts the money and finds that
his money is worth $4.50 and that he has 24 coins in his pocket. How many of each type of coin does he
have?
w w
l
l
3
Do # 12, p 402 text in your homework booklet.
E.g.: A.J. wrote a problem about the number of senior high students attending the last dance and the
number of junior high students who attended the same dance. He modeled the situation with the
following linear system
5 3 270
70
s j
s j
What situation might the equations represent? What does each variable represent?
Situation: Admission to the dance cost $5 for senior high students and $3 for junior high students. The
total amount of money collected was $270, and 70 people attended the dance. How many senior high
students and how many junior high students attended the dance?
Let s represent the number of senior high students attending the dance.
Let j represent the number of junior high students attending the dance.
Do #’s 6, 13, pp. 402-403 text in your homework booklet.
4
§7.2 & 7.3 Solving a System of Linear Functions Graphically (1 class)
Read Lesson Focus p. 403 & Lesson Focus p. 411 text.
Outcomes
1. Determine and verify the solution of a system of linear equations graphically, with and
without technology. pp. 405-408 & pp. 412-413.
2. Explain the meaning of the point of intersection of the graphs of a system of linear equations.
p. 404
E.g.: Solve 3 7
5 9
x y
x y
by carefully graphing each line.
Step 1: Rewrite each equation in slope-intercept form.
Step 2: Carefully graph each line on the same axes.
Step 3: Determine the coordinates of the point of
intersection of the two lines.
The point of intersection has coordinates 2,1 .
Step 4: Write the solution.
The solution is 2 and 1x y
Step 5: Check the solution in the original equations.
. . . 3 2 1 6 1 7 . .L H S R H S
. . . 5 2 1 10 1 9 . . .L H S R H S
*****The solution to a system of equations can be found from the point of __________________ on
the graph.*****
3 7
3 3 7 3
7 3
3 7
3; 7
1
x y
x x y x
y x
y x
m b
5 9
5 5 9 5
9 5
5 9
5; 9
1
x y
x x y x
y x
y x
m b
5
E.g.: Solve 2 3 20
7 2 5
x y
x y
by carefully graphing each line.
Step 1: Rewrite each equation in slope-intercept form.
Step 2: Carefully graph each line on the same axes.
Step 3: Determine the coordinates of the point of
intersection of the two lines.
The point of intersection has coordinates 1,6 .
Step 4: Write the solution.
The solution is 1 and 6x y
Step 5: Check the solution in the original equations.
. . . 2 1 3 6 2 18 20 . .L H S R H S
. . . 7 1 2 6 7 12 5 . . .L H S R H S
Do #’s 3, 4, 5 a i), iii), iv), 6, p. 409 text in your homework booklet.
Solving a Linear System Using Graphing Technology
In #4 b, p. 409, you could only give an approximate the solution to the linear system. If you use
graphing technology, your solution can be much more accurate and precise and the graphing much
easier.
2 3 20
2 2 3 2 20
3 2 20
3 2 20
3 3 3
2 20
3 3
2 20; 6.6
3 3
x y
x x y x
y x
y x
y x
m b
7 2 5
7 7 2 5 7
2 7 5
2 7 5
2 2 2
7 5
2 2
7 5; 2.5
2 2
x y
x x y x
y x
y x
y x
m b
6
E.g.: Solve 2 5
12 8 75
x y
x y
by graphing each line.
Step 1: Rewrite each equation in slope-intercept form.
Step 2: Graph each line on the same axes using technology.
Step 3: Determine the coordinates of the point of intersection of the two lines.
The point of intersection has coordinates 1.25, 7.5 .
Step 4: Write the solution.
The solution is 1.25 and 7.5x y
Step 5: Check the solution in the original equations.
. . . 2 1.25 7.5 2.5 7.5 5 . .L H S R H S
. . . 12 1.25 8 7.5 15 60 75 . . .L H S R H S
Do # 7, p. 409 text in your homework booklet.
2 5
2 2 5 2
2 5
x y
x x y x
y x
12 8 75
12 12 8 75 12
8 12 75
8 8 8
3 75
2 8
x y
x x y x
y x
y x
7
Problem Solving Using Linear Systems
E.g.: Two companies embroider logos on ball caps. Company A charges
$180 for set up and $4 per cap. Company B charges $256 for set up and $3
per cap. A linear system which models this situation is
180 4
256 3
C n
C n
where C represents the total cost and n represents the number of ball caps embroidered.
a) Graph the linear system.
b) Use the graph to determine:
i. the number of ball caps that must be embroider for each company to cost the same
amount.
ii. the number of ball caps for which Company B is cheaper
a) Graph 180 4
256 3
y x
y x
on your G.C. and adjust the window until you can see the point of
intersection of the two lines.
b)
Both companies cost the same for 76 caps. The cost for both companies would be $484.
c) The thick line lies below the thin line when the number of ball caps is greater than 76. So
company B would be cheaper if you had more than 76 caps embroidered.
Do #’s 8, 9, 12, 13, 15, pp. 409-410 text in your homework booklet.
8
§7.4 Using the Substitution Method to Solve a System of Linear Equations (2 classes)
Read Lesson Focus p. 416 text.
Outcomes
1. Determine and verify the solution of a system of linear equations algebraically. p. 417
2. Explain a strategy to solve a system of linear equations. p. 417
3. Solve a problem that involves a system of linear equations. p. 420
Solving a Linear System Using Substitution
E.g.: Solve 2 5
12 8 75
x y
x y
using substitution.
Step 1: Rearrange one of the equations to solve for x or y, whichever is easier. Usually this will be a
variable with coefficient of 1.
We will rearrange the first equation and solve for y because it has a coefficient of 1.
Step 2: Substitute into the OTHER equation.
In the second equation, we will replace y by 5 2x .
12 8 5 2 75x x
Step 3: Solve the equation with only one variable.
We will solve 12 8 5 2 75x x for x.
12 8 5 2 75
12 40 16 75
28 40 75
28 40 40 75 40
28 35
28 35
28 28
5 or 1.25
4
x x
x x
x
x
x
x
x
2 5
2 2 5 2
5 2
x y
x x y x
y x
9
Step 4: Substitute the value just found into the equation in Step 1.
We will replace x in 5 2y x by 1.25.
5 2 1.25 5 2.5 7.5y
Step 5: Write the solution.
The solution is 1.25 and 7.5x y
Step 6: Check the solution in the original equations.
. . . 2 1.25 7.5 2.5 7.5 5 . .L H S R H S
. . . 12 1.25 8 7.5 15 60 75 . . .L H S R H S
E.g.: Solve 2 13
4 3 1
x y
x y
using substitution.
Step 1: Solve for y in the first equation.
2 13
2 2 13 2
13 2
13 2
1 1 1
13 2
x y
x x y x
y x
y x
y x
Step 2: Substitute into the OTHER equation and solve for x.
4 3 13 2 1
4 39 6 1
10 39 1
10 39 39 1 39
10 40
10 40
10 10
4
x x
x x
x
x
x
x
x
10
Step 3: Substitute the value for x obtained in the last step into 13 2y x .
13 2 4
13 8
5
y
y
y
Step 4: Write the solution.
The solution is 4 and 5x y
Step 5: Check the solution in the original equations.
. . . 2 4 5 8 5 13 . .L H S R H S
. . . 4 4 3 5 16 15 1 . . .L H S R H S
E.g.: Solve 6 9
3 2 23
x y
x y
using substitution.
Step 1: Solve for x in the first equation.
6 9
6 6 9 6
9 6
x y
x y y y
x y
Step 2: Substitute into the OTHER equation and solve for y.
3 9 6 2 23
27 18 2 23
27 20 23
27 27 20 23 27
20 50
20 50
20 20
5 or 2.5
2
y y
y y
y
y
y
y
y
Step 3: Substitute the value for y obtained in the last step into 9 6x y .
9 6 2.5
9 15
6
x
x
x
11
Step 4: Write the solution.
The solution is 6 and 2.5x y
Step 5: Check the solution in the original equations.
. . . 6 6 2.5 6 15 9 . .L H S R H S
. . . 3 6 2 2.5 18 5 23 . . .L H S R H S
E.g.: Below is Kristen’s solution to 3 5
5 2 8
m n
m n
. Identify and correct any error(s) she made.
Step 1 Step 2 Step 3 Step 4
3 5
3 3 5 3
5 3
m n
m m n m
n m
5 2 5 3 8
5 10 3 8
2 10 8
2 10 10 8 10
2 18
9
m m
m m
m
m
m
m
The solution is
9 and 22m n
Kristen did not solve for n properly in step 1. She did not include -1 as the coefficient for n.
She also did not do the distributive property correctly in step 2. She did not multiply -2 by -3m.
Step 1 Step 2 Step 3 Step 4
3 5
3 3 5 3
5 3
5 3
1 1 1
5 3
m n
m m n m
n m
n m
n m
5 2 5 3 8
5 10 6 8
10 8
10 10 8 10
2
2
m m
m m
m
m
m
m
The solution is
2 and 1m n
Do # 5, p. 425 text in your homework booklet.
Solving a Linear System Containing Fractions Using Substitution
E.g.: Solve
1 11
6 3
1 21
4 3
a b
a b
using substitution.
5 3 9
5 27
22
n
n
n
5 3 2
5 6
1
n
n
n
12
Step 1: Use the LCM of the denominators in each equation to get rid of the fractions.
In the first equation, the LCM of the denominators is 6, so we multiply each term by 6.
6 1 6 1
6 11 6 1 3
2 6
a b
a b
In the second equation, the LCM of the denominators is 12, so we multiply each term by 12.
12 1 12 2
12 11 4 1 3
3 8 12
a b
a b
We will now solve 2 6
3 8 12
a b
a b
instead of
1 11
6 3
1 21
4 3
a b
a b
.
Step 2: Rearrange one of the equations to solve for a or b, whichever is easier. Usually this will be a
variable with coefficient of 1.
We will rearrange the first equation and solve for a because it has a coefficient of 1.
Step 3: Substitute into the OTHER equation and solve for the variable.
In the second equation, we will replace a by 6 2b .
3 6 2 8 12
18 6 8 12
18 2 12
18 18 2 12 18
2 30
2 30
2 2
15
b b
b b
b
b
b
b
b
2 6
2 2 6 2
6 2
a b
a b b b
a b
13
Step 4: Substitute the value just found into the equation in Step 2.
We will replace b in 6 2a b by -15.
6 2 15 6 30 36a
Step 5: Write the solution.
The solution is 36 and 15a b
Step 6: Check the solution in the original equations.
1 1 36 15
. . . 36 15 6 5 1 . .6 3 6 3
L H S R H S
1 2 36 30
. . . 36 15 9 10 1 . .4 3 4 3
L H S R H S
E.g.: Solve
1 11
2 3
12
4
x y
x y
using substitution.
Step 1: Use the LCM of the denominators in each equation to get rid of the fractions.
In the first equation, the LCM of the denominators is 6, so we multiply each term by 6.
6 1 6 1
6 11 2 1 3
3 2 6
x y
x y
In the second equation, the LCM of the denominators is 4, so we multiply each term by 4.
4 1
4 4 21 4
4 8
x y
x y
We will now solve 3 2 6
4 8
x y
x y
instead of
1 11
2 3
12
4
x y
x y
.
14
Step 2: Rearrange one of the equations to solve for x or y, whichever is easier. Usually this will be a
variable with coefficient of 1.
We will rearrange the second equation and solve for y because it has a coefficient of 1.
Step 3: Substitute into the OTHER equation and solve for the variable.
In the first equation, we will replace y by 8 4x .
3 2 8 4 6
3 16 8 6
11 16 6
11 16 16 6 16
11 22
11 22
11 11
2
x x
x x
x
x
x
x
x
Step 4: Substitute the value just found into the equation in Step 2.
We will replace x in 8 4y x by 2.
8 4 2
8 8
0
y
y
y
Step 5: Write the solution.
The solution is 2 and 0x y
Step 6: Check the solution in the original equations.
1 1 2 0
. . . 2 0 1 0 1 . .2 3 2 3
L H S R H S
1
. . . 2 0 2 0 2 . .4
L H S R H S
Do #’s 8, 19, pp. 425-426 text in your homework booklet.
4 8
4 4 8 4
8 4
x y
x x y x
y x
15
Problem Solving Using Linear Systems
E.g.: The perimeter of a rectangle is 46cm. What are the dimensions of the rectangle if the length is 4
less than twice the width?
Let l represent the length of the rectangle. Let w represent the width of the rectangle.
Since the perimeter is 46, n2 2 46 Eq 1l w
Since the length is 4 less than twice the width, we can write the equation:
n2 4 Eq 2l w
We can combine equations 1 and 2 to make a linear system 2 2 46
2 4
l w
l w
Substitute 2 4w for l in the first equation.
2 2 4 2 46
4 8 2 46
6 8 46
6 8 8 46 8
6 54
9cm
w w
w w
w
w
w
w
If 9cmw then 2 9 4 18 4 14cml
The dimensions of the rectangle are 9cm by 14cm.
E.g.: 170 junior and senior high students were surveyed to determine their cell phone use. 128 students
reported heavy cell phone use. This was 80% of the junior high students and 70% of the senior high
students. How many senior high and how many junior high students were in the study?
Let j represent the number of junior high students in the study.
Let s represent the number of senior high students in the study.
The system of equations that models this situation is 170
0.80 0.70 128
j s
j s
Using the first equation, 170j s
Substituting into the second equation gives 0.80 170 0.70 128s s .
16
Solving this equation gives
0.80 170 0.70 128
136 0.80 0.70 128
136 0.10 128
136 136 0.10 128 136
0.10 8
0.10 8
0.10 0.10
80
s s
s s
s
s
s
s
s
If 80s , then 170 80 90j
There were 80 senior and 90 junior high students in the survey.
Do #’s 11, 12, 15, pp. 425-426 text in your homework booklet.
17
§7.5 Using the Elimination Method to Solve a System of Linear Equations (2 classes)
Read Lesson Focus p. 428 text.
Outcomes
1. Determine and verify the solution of a system of linear equations algebraically. p. 417
2. Explain a strategy to solve a system of linear equations. p. 417
3. Solve a problem that involves a system of linear equations. p. 420
Solving a Linear System Using Elimination
E.g.: Solve 2 5
12 8 75
x y
x y
using elimination.
Step 1: Multiply one or both of the equations by a constant so that the coefficients of the x terms OR the
coefficients of the y terms are opposites.
We will multiply the first equation by 8 so that the first equation has an 8y term and the second equation
has a -8y term.
16 8 408 2 5
12 8 7512 8 75
x yx y
x yx y
Step 2: Add the equations to eliminate one of the variables.
We will add the equations to eliminate the y term.
Step 3: Solve the equation in step 2.
28 35
28 35
28 28
51.25
4
x
x
x
Step 4: Substitute the value from step 3 into EITHER of the original equations and solve.
We will replace x by 1.25 in the first equation.
2 1.25 5
2.5 5
2.5 2.5 5 2.5
7.5
y
y
y
y
16 8 40
12 8 75
28 0 35 28 35
x y
x y
x y x
18
Step 5: Write the solution.
The solution is 1.25 and 7.5x y
Step 6: Check the solution in the original equations.
. . . 2 1.25 7.5 2.5 7.5 5 . .L H S R H S
. . . 12 1.25 8 7.5 15 60 75 . . .L H S R H S
E.g.: Solve 2 13
4 3 1
x y
x y
using elimination.
Step 1: Multiply one or both of the equations by a constant so that the coefficients of the x terms OR the
coefficients of the y terms are opposites.
We will multiply the first equation by 3 so that the first equation has an -3y term and the second
equation has a 3y term.
6 3 393 2 13
4 3 14 3 1
x yx y
x yx y
Step 2: Add the equations to eliminate one of the variables.
We will add the equations to eliminate the y term.
Step 3: Solve the equation in step 2.
10 40
10 40
10 10
4
x
x
x
6 3 39
4 3 1
10 0 40 10 40
x y
x y
x y x
19
Step 4: Substitute the value from step 3 into EITHER of the original equations and solve.
We will replace x by 4 in the first equation.
2 4 13
8 13
8 8 13 8
5
5
1 1
5
y
y
y
y
y
y
Step 5: Write the solution.
The solution is 4 and 5x y
Step 6: Check the solution in the original equations.
. . . 2 4 5 8 5 13 . .L H S R H S
. . . 4 4 3 5 16 15 1 . . .L H S R H S
E.g.: Solve 6 9
3 2 23
x y
x y
using elimination.
Step 1: Multiply one or both of the equations by a constant so that the coefficients of the x terms OR the
coefficients of the y terms are opposites.
We will multiply the second equation by 3 so that the first equation has an 6y term and the second
equation has a -6y term.
6 9 6 9
3 3 2 23 9 6 69
x y x y
x y x y
Step 2: Add the equations to eliminate one of the variables.
We will add the equations to eliminate the y term.
6 9
9 6 69
10 0 60 10 60
x y
x y
x y x
20
Step 3: Solve the equation in step 2.
10 60
10 60
10 10
6
x
x
x
Step 4: Substitute the value from step 3 into EITHER of the original equations and solve.
We will replace x by -6 in the first equation.
6 6 9
6 6 6 9 6
6 15
6 15
6 6
52.5
2
y
y
y
y
y
Step 5: Write the solution.
The solution is 6 and 2.5x y
Step 6: Check the solution in the original equations.
. . . 6 6 2.5 6 15 9 . .L H S R H S
. . . 3 6 2 2.5 18 5 23 . . .L H S R H S
E.g.: Solve 3 4 18
2 3 5
s t
s t
using elimination.
Step 1: Multiply the first equation by 3 and the second equation by 4.
3 3 4 18 9 12 54
4 2 3 5 8 12 20
s t s t
s t s t
Step 2: We will add the equations to eliminate the y term.
9 12 54
8 12 20
17 0 34 17 34
s t
s t
s t s
21
Step 3: Solve the equation in step 2.
17 34
17 34
17 17
2
s
s
s
Step 4: Replace s by 2 in the first equation.
3 2 4 18
6 4 18
6 6 4 18 6
4 12
4 12
4 4
3
t
t
t
t
t
t
Step 5: Write the solution.
The solution is 2 and 3s t
Step 6: Check the solution in the original equations.
. . . 3 2 4 3 6 12 18 . .L H S R H S
. . . 2 2 3 3 4 9 5 . . .L H S R H S
Do #’s 20, 3 a, b, 6, 7 b, d, pp. 437-438 text in your homework booklet.
Solving a Linear System Containing Fractions Using Elimination
E.g.: Solve
2 12
3 5
1 17
3 2
x y
x y
using elimination.
Step 1: Multiply the first equation by 15 and the second equation by 6.
2 115 2
10 3 303 5
2 3 421 16 7
3 2
x yx y
x yx y
22
Step 2: Add the equations to eliminate the y term.
10 3 30
2 3 42
12 0 12 12 12
x y
x y
x y x
Step 3: Solve the equation in step 2.
12 12
12 12
12 12
1
x
x
x
Step 4: Replace x by -1 in the second equation and solve.
1 1
1 73 2
1 17
3 2
1 16 7
3 2
2 3 42
2 2 3 42 2
3 40
3 40
3 3
40
3
y
y
y
y
y
y
y
y
Step 5: Write the solution.
The solution is 40
1 and 3
x y
Step 6: Check the solution in the original equations.
2 1 40 2 40 10 40 30
. . . 1 2 . .3 5 3 3 15 15 15 15
L H S R H S
1 1 40 1 40 1 20 21
. . . 1 7 . . .3 2 3 3 6 3 3 3
L H S R H S
Do # 12, p. 438 text in your homework booklet.
23
Problem Solving Using Linear Systems
E.g.: An artist was commissioned to make a 625g statue of an eagle out of a metal alloy that is 40%
silver. She has some alloy that is 50% silver and some more alloy that is 25% silver. How much of each
alloy should she mix to get the 40% silver alloy that she needs?
Let f represent the amount of the 50% silver alloy.
Let t represent the amount of the 25% silver alloy.
The linear system that models this situation is
625 625
0.5 0.25 0.40 625 0.5 0.25 250
f t f t
f t f t
.
Solving for f in the first equation gives 625f t .
Substituting for f in the second equation and solving gives
0.5 625 0.25 250
312.5 0.5 0.25 250
312.5 0.25 250
312.5 312.5 0.25 250 312.5
0.25 62.5
0.25 62.5
0.25 0.25
250g
t t
t t
t
t
t
t
t
If 250t then,
250 625
250 250 625 250
375g
f
f
f
The artist needs 375g of 25% silver alloy and 375g of 50% silver alloy.
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E.g.: A butcher has supplies of lean beef containing 15% fat, and fat trim which is 100% fat. How many
kilograms of each, to the nearest tenth of a kg, would be needed to make 50kg of hamburger which is
25% fat? [44.1kg & 5.9kg]
E.g.: A vinegar-water solution is used to wash windows. If 1200L of a 28% vinegar solution are
required, how much of 16% and 36% vinegar solutions should be mixed? [480L,720L]
Do #’s 16, 14, 17 p. 438 text in your homework booklet.
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§7.6 Properties of Systems of Linear Equations (1 class)
Read Lesson Focus p. 442 text.
Outcomes
1. Define and draw coincident lines. p. 443
2. Explain what is meant by an infinite number of solutions. p. 443
3. Explain, using examples, why a system of equations may have no solution, one solution or an
infinite number of solutions. pp. 444-447
4. By examining the coefficients of the equations of a linear system determine if the system has
one solution, no solution, or an infinite number of solutions. pp. 444-447
Solving a Linear System Using Substitution
E.g.: Compare the coefficients of the terms of 10 15 18
2 3 5
s t
s t
. What do you notice?
Solve 10 15 18
2 3 5
s t
s t
using any method you wish.
Let’s solve by graphing. First we must solve each equation for s or t. We will solve each equation for s.
10 18 15 2 5 3
10 18 15 2 5 3
10 10 10 2 2 2
2.5 1.51.8 1.5
s t s t
s t s t
s ts t
Note that the slope of both lines is 1.5. This means that the
lines are parallel and will never intersect (see graph below).
This means there is no solution to this linear system.
Could we have determined that there was no solution just by looking at the equations in the linear
system? If we multiply the second equation by 5, we get 10 15 25s t . You should realize that for a
single value of s and a single value of t, it is impossible for 10 15 25s t and for 10 15 18s t .
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E.g.: Compare the coefficients of the terms of 10 15 25
2 3 5
s t
s t
. What do you notice?
Solve 10 15 25
2 3 5
s t
s t
using any method you wish.
Let’s solve by graphing. First we must solve each equation for s or t. We will solve each equation for s.
10 25 15 2 5 3
10 25 15 2 5 3
10 10 10 2 2 2
2.5 1.52.5 1.5
s t s t
s t s t
s ts t
Note that both equations are identical. This means that the lines
are coincident (their graphs are coincide) (see graph to the
right). This means there are an infinite number (without bound)
number of solutions to this linear system.
Could we have determined that there were an infinite number of solutions just by looking at the
equations in the linear system? If we multiply the second equation by 5, we get the first equation. You
should realize that any solution for 10 15 25s t is also a solution for 2 3 5s t .
Summary
When we solve linear systems, there are three possibilities:
