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Linear Systems
• Systems of Linear Equations
• Solving Systems of Equations by
Substitution
1
Systems of Equations
A set of equations is called a system of
equations.
The solutions must satisfy each equation in the
system.
If all equations in a system are linear, the system
is a system of linear equations, or a linear
system.
Systems of Linear Equations:
A solution to a system of equations is an
ordered pair that satisfy all the equations in
the system.
A system of linear equations can have:
1. Exactly one solution
2. No solutions
3. Infinitely many solutions
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Systems of Linear Equations:
There are four ways to solve systems of linear
equations:
1. By graphing
2. By substitution
3. By addition (also called elimination)
4. By multiplication
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Solving Systems by Graphing:
When solving a system by graphing:
1. Find ordered pairs that satisfy each of the
equations.
2. Plot the ordered pairs and sketch the
graphs of both equations on the same axis.
3. The coordinates of the point or points of
intersection of the graphs are the solution or
solutions to the system of equations.
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Solving Systems by Graphing:
Consistent
Inconsistent
One solution
No solution
Lines intersect
Lines are parallel
Dependent
Infinite number of
solutions
Coincide-Same
line
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Linear System in Two Variables
Three possible solutions to a linear system in two
variables:
One solution: coordinates of a point
No solutions: inconsistent case
Infinitely many solutions: dependent case
2x – y = 2
x + y = -2
2x – y = 2
-y = -2x + 2
y = 2x – 2
x + y = -2
y = -x - 2
Different slope, different intercept!
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3x + 2y = 3
3x + 2y = -4
3x + 2y = 3
2y = -3x + 3
y = -3/2 x + 3/2
3x + 2y = -4
2y = -3x -4
y = -3/2 x - 2
Same slope, different intercept!!
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x – y = -3
2x – 2y = -6
x – y = -3
-y = -x – 3
y =x+3
2x – 2y = -6
-2y = -2x – 6
y=x+3
Same slope, same intercept!
Same equation!!
Determine Without Graphing:
• There is a somewhat shortened way to
determine what type (one solution, no
solutions, infinitely many solutions) of
solution exists within a system.
• Notice we are not finding the solution, just
what type of solution.
• Write the equations in slope-intercept form:
y = mx + b.
(i.e., solve the equations for y, remember
that m = slope, b = y - intercept).
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Determine Without Graphing:
Once the equations are in slope-intercept form,
compare the slopes and intercepts.
One solution – the lines will have different slopes.
No solution – the lines will have the same slope,
but different intercepts.
Infinitely many solutions – the lines will have the
same slope and the same intercept.
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Determine Without Graphing:
Given the following lines, determine what type
of solution exists, without graphing.
Equation 1:
3x = 6y + 5
Equation 2:
y = (1/2)x – 3
Writing each in slope-intercept form (solve for y)
Equation 1:
y = (1/2)x – 5/6
Equation 2:
y = (1/2)x – 3
Since the lines have the same slope but
different y-intercepts, there is no solution to the
system of equations. The lines are parallel.
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Substitution Method:
Procedure for Substitution Method
1. Solve one of the equations for one of the variables.
2. Substitute the expression found in step 1 into the
other equation.
3. Now solve for the remaining variable.
4. Substitute the value from step 2 into the equation
written in step 1, and solve for the remaining
variable.
Substitution Method:
1. Solve the following system of equations by
substitution.
y  x3
Step 1 is already completed.
x  y  5
Step 2:Substitute x+3 into
2nd equation and solve.
x  ( x  3)  5
2x  3  5
2x  8
x  4
Step 3: Substitute –4 into 1st
equation and solve.
y  x3
y  4  3
y  1
The answer: ( -4 , -1)
1) Solve the system using substitution
x+y=5
y=3+x
Step 1: Solve an
equation for one
variable.
Step 2: Substitute
Step 3: Solve the
equation.
The second equation is
already solved for y!
x+y=5
x + (3 + x) = 5
2x + 3 = 5
2x = 2
x=1
1) Solve the system using substitution
x+y=5
y=3+x
Step 4: Plug back in to
find the other
variable.
Step 5: Check your
solution.
x+y=5
(1) + y = 5
y=4
(1, 4)
(1) + (4) = 5
(4) = 3 + (1)
The solution is (1, 4). What do you think the answer
would be if you graphed the two equations?
2) Solve the system using substitution
3y + x = 7
4x – 2y = 0
Step 1: Solve an
equation for one
variable.
It is easiest to solve the
first equation for x.
3y + x = 7
-3y
-3y
x = -3y + 7
Step 2: Substitute
4x – 2y = 0
4(-3y + 7) – 2y = 0
2) Solve the system using substitution
3y + x = 7
4x – 2y = 0
Step 3: Solve the
equation.
Step 4: Plug back in to
find the other
variable.
-12y + 28 – 2y = 0
-14y + 28 = 0
-14y = -28
y=2
4x – 2y = 0
4x – 2(2) = 0
4x – 4 = 0
4x = 4
x=1
2) Solve the system using substitution
3y + x = 7
4x – 2y = 0
Step 5: Check your
solution.
(1, 2)
3(2) + (1) = 7
4(1) – 2(2) = 0
Deciding whether an ordered pair is a
solution of a linear system.
The solution set of a linear system of equations contains all ordered
pairs that satisfy all the equations at the same time.
Example 1: Is the ordered pair a solution of the given system?
2x + y = -6
Substitute the ordered pair into each equation.
x + 3y = 2
Both equations must be satisfied.
A) (-4, 2)
B) (3, -12)
2(-4) + 2 = -6
2(3) + (-12) = -6
(-4) + 3(2) = 2
(3) + 3(-12) = 2
-6 = -6
-6 = -6
2=2
-33  -6
 Yes
 No
Substitution Method
Example Solve the system.
3x  2 y  11
x  y  3
Solution y  x  3
3 x  2( x  3)  11
3 x  2 x  6  11
5x  5
x 1
y  1 3
y4
(1)
(2)
Solve (2) for y.
Substitute y = x + 3 in (1).
Solve for x.
Substitute x = 1 in y = x + 3.
Solution set: {(1, 4)}
Systems of Linear Equations in Two Variables
Solving Linear Systems by Graphing.
One way to find the solution set of a linear system of equations is to graph
each equation and find the point where the graphs intersect.
Example 1: Solve the system of equations by
graphing.
A) x + y = 5
B) 2x + y = -5
2x - y = 4
-x + 3y = 6
Solution: {(3,2)}
Solution: {(-3,1)}
Systems of Linear Equations in Two Variables
Solving Linear Systems by Graphing.
There are three possible solutions to a system of linear equations in two
variables that have been graphed:
1) The two graphs intersect at a single point. The coordinates give the solution of
the system. In this case, the solution is “consistent” and the equations are
“independent”.
2) The graphs are parallel lines. (Slopes are equal) In this case the system is
“inconsistent” and the solution set is 0 or null.
3) The graphs are the same line. (Slopes and y-intercepts are the same) In this
case, the equations are “dependent” and the solution set is an infinite set of
ordered pairs.
4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by
Method of Substitution.
Step 1: Solve one of the equations for either variable
Step 2: Substitute for that variable in the other equation
(The result should be an equation with just one variable)
Step 3: Solve the equation from step 2
Step 4: Substitute the result of Step 3 into either of the original
equations and solve for the other value.
Step 6: Check the solution and write the solution set.
4-1 Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by
Method of Substitution.
Example 6: Solve the system :
4x + y = 5
2x - 3y =13
Step 1: Choose the variable y to solve for in the top equation:
y = -4x + 5
Step 2: Substitute this variable into the bottom equation
2x - 3(-4x + 5) = 13
2x + 12x - 15 = 13
Step 3: Solve the equation formed in step 2
14x = 28
x=2
Step 4: Substitute the result of Step 3 into either of the original equations and solve for the other
value. 4(2) + y = 5
y = -3
Solution Set: {(2,-3)}
Step 5: Check the solution and write the solution set.
Systems of Linear Equations in Two Variables
Solving Linear Systems of two variables by
Method of Substitution.
Example 7:
Solve the system : 1
1
1
1
1
1
y
rewrite as  4[ x  y  ]  2 x  y  2
2
4
2
2
4
2
2 x  5 y  22
x
Solve :
2x  y  2
-2x  5 y  22
y = -2x + 2
-2x + 5(-2x + 2) = 22
-12x = 12
x = -1
Solution Set: {(-1,4)}
-2x - 10x + 10 = 22
2(-1) + y = 2
y=4
Your Turn:
3x – y = 4
x = 4y - 17
Your Turn:
2x + 4y = 4
3x + 2y = 22
Clearing Fractions or Decimals



Systems without a Single Point Solution
0 = 4 untrue
Inconsistent Systems - how can you tell?
An inconsistent system
has no solutions.
(parallel lines)
Substitution Technique
( A)
( B)



y  3 x  5
y  3 x  2
 3 x  5  3 x  2
 3x
 3x
5
2
2
2
7  0 inconsiste nt
0 = 0 or n = n
Dependent Systems – how can you tell?
A dependent system has
infinitely many solutions.
(it’s the same line!)
Substitution Technique



( A)
3y  2x  6
( B)  12 y  8 x  24
( B) 8 x  12 y  24
x  32 y  3
( A) 3 y  2( 32 y  3)  6
3y  3y  6  6
6  6 dependent
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Modeling Examples:
The reason to learn about systems of equations
is to learn how to solve real world problems.
Study Example 8 on page 360 in the text.
Notice how the original equations are set up
based on the data in the question.
Also note that we are trying to determine when
the total cost at each garage will be the same.
To do this, set the two cost equations equal to
each other and solve. You will see this type of
problem often.
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Modeling Examples:
Study Example 9 on page 361 in the text. This
is a mixture problem. Notice how the original
equations are set up based on the data in the
question.
Once the equations are set up, the 2nd equation
is multiplied by 100 to remove the decimal.
This is a common occurrence, so make sure
you know how to do this.
Note: The example is solved using the addition
method. It can also be solved by substitution.
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Modeling Examples:
4. Read problem 40 on page 362 of the text –
“basketball game”.
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Modeling Examples:
4. Read problem 40 on page 362 of the text –
“basketball game”.
First assign the variables:
let x = # of 2 point shots
let y = # of 3 point shots
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Modeling Examples:
4. Read problem 40 on page 362 of the text –
“basketball game”.
First assign the variables:
let x = # of 2 point shots
let y = # of 3 point shots
Writing the 1st equation:
They made 45 goals in a recent game
x + y = 45
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Modeling Examples:
4 continued.
Writing the 2nd equation:
40
Modeling Examples:
4 continued.
Writing the 2nd equation:
Some 2 pointers, some 3 pointers, for a total
score of 101 points
2x + 3y = 101
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Modeling Examples:
4 continued.
Writing the 2nd equation:
Some 2 pointers, some 3 pointers, for a total
score of 101 points
2x + 3y = 101
In words, the equation says 2 times the number
of 2 point shots plus 3 times the number of 3
point shots totals 101 points.
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Modeling Examples:
4 continued.
The two equations are:
x + y = 45
2x + 3y = 101
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Modeling Examples:
4 continued.
The two equations are:
-2( x + y = 45 ) Lets eliminate x, multiply the
st equation by –2.
entire
1
2x + 3y = 101
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Modeling Examples:
4 continued.
The two equations are:
-2( x + y = 45 ) Lets eliminate x, multiply the
st equation by –2.
entire
1
2x + 3y = 101
-2x + -2y = -90
2x + 3y = 101
45
Modeling Examples:
4 continued.
The two equations are:
-2( x + y = 45 )
2x + 3y = 101
-2x + -2y = -90
2x + 3y = 101
y = 11
Add down to eliminate x.
Substitute y into the 1st
equation. x + 11 = 45, so
x = 34.
34 - 2 point shots and
11 - 3 point shots.
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Modeling Examples:
5. Read problem 44 on page 363 in the text –
A Milk Mixture.
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Modeling Examples:
5. Read problem 44 on page 363 in the text –
A Milk Mixture.
First assign the variables:
let x = # gallons of 5% milk
let y = # gallons of skim (0%) milk
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Modeling Examples:
5. Read problem 44 on page 363 in the text –
A Milk Mixture.
First assign the variables:
let x = # gallons of 5% milk
let y = # gallons of skim (0%) milk
Writing the 1st equation:
x + y = 100
This is because they want to make a mixture
totaling 100 gallons of milk.
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Modeling Examples:
5. Continued
Writing the 2nd equation:
50
Modeling Examples:
5. Continued
Writing the 2nd equation:
0.05x + 0.0y = 0.035(100)
Basically, we are multiplying the 1st equation by
the percent butterfat of the milk. Our final
mixture should be 3.5%, so we multiply
0.035(100), since we want 100 total gallons.
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Modeling Examples:
5. Continued
The two equations are:
x+
y = 100
0.05x + 0.0y = 0.035(100)
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Modeling Examples:
5. Continued
The two equations are:
x+
y = 100
0.05x + 0.0y = 0.035(100)
Next, multiply the 2nd equation by 1000 to remove
the decimal. This gives us the following system
of equations:
x + y = 100
50x + 0y = 35(100)
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Modeling Examples:
5. Continued
The two equations are:
x+
y = 100
0.05x + 0.0y = 0.035(100)
Next, multiply the 2nd equation by 1000 to remove
the decimal. This gives us the following system
of equations:
x + y = 100
50x + 0y = 35(100)
Solve the system (use substitution since the 2nd
equation has only one variable). The answer
follows on the next slide.
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Modeling Examples:
5. Continued
The answer is 70 gallons of 5% milk and 30
gallons of skim (0%) milk.
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Modeling Examples:
6. Read problem 48 on page 363 in the text –
School Play Tickets.
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Modeling Examples:
6. Read problem 48 on page 363 in the text –
School Play Tickets.
First assign the variables:
let x = # of adult tickets sold ($5 per ticket)
let y = # of student tickets sold ($2 per ticket)
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Modeling Examples:
6. Read problem 48 on page 363 in the text –
School Play Tickets.
First assign the variables:
let x = # of adult tickets sold ($5 per ticket)
let y = # of student tickets sold ($2 per ticket)
Writing the 1st equation:
x + y = 250
Since a total of 250 tickets were sold.
58
Modeling Examples:
6. Continued
Writing the 2nd equation:
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Modeling Examples:
6. Continued
Writing the 2nd equation:
5x + 2y = 950
Basically, we multiplied the 1st equation by the
price of the tickets, and set it equal to the
amount of money collected.
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Modeling Examples:
6. Continued
Writing the 2nd equation:
5x + 2y = 950
Basically, we multiplied the 1st equation by the
price of the tickets, and set it equal to the
amount of money collected.
Do you see how this is similar to example #4?
The 2 and 3 point shots?
61
Modeling Examples:
6. Continued
The two equations are:
x + y = 250
5x + 2y = 950
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Modeling Examples:
6. Continued
The two equations are:
x + y = 250
5x + 2y = 950
Can you solve the system using either
substitution or addition? The answer follows on
the next slide.
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Modeling Examples:
6. Continued
The answer is 150 adult tickets were sold, and
100 student tickets were sold.
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