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Chapter 3
Systems of Linear Equations
§ 3.1
Systems of Linear Equations in Two Variables
Systems of Equations
EXAMPLE
Determine whether (3,2) is a solution of the system
 3 x  6 y  10
5 x  8 y  2
SOLUTION
Because 3 is the x-coordinate and 2 is the y-coordinate of the
point(3,2), we replace x with 3 and y with 2.
?
 3x  6 y 
10
 33  62  10
?
?
 9  12  10
3  10 false
Since the result is false, (3,2) is
NOT a solution for the system.
Also, I need not check the other
equation since the first one failed.
Blitzer, Intermediate Algebra, 4e – Slide #3
Systems of Equations
Solve Systems of Two Linear Equations in Two
Variables, x and y, by Graphing
1) Graph the first equation.
2) Graph the second equation on the same set of axes.
3) If the lines representing the two graphs intersect at a point, determine the
coordinates of this point of intersection. The ordered pair is the solution to the
system.
4) Check the solution in both equations.
NOTE: In order for this method to be useful, you must graph the lines very accurately.
Blitzer, Intermediate Algebra, 4e – Slide #4
Systems of Equations
EXAMPLE
Solve by graphing:
4x  y  4
3x  y  3
SOLUTION
1) Graph the first equation. I first rewrite the equation in
slope-intercept form.
4x  y  4

y  4 x  4
m = -4 = -4/1, b = 4
Now I can graph the equation.
Blitzer, Intermediate Algebra, 4e – Slide #5

Systems of Equations
CONTINUED
2) Graph the second equation on the same set of axes.
I first rewrite the equation in slope-intercept form.
3x  y  3
 y  3 x  3
y  3x  3

m = 3 = 3/1, b = -3
Now I can graph the equation.
Blitzer, Intermediate Algebra, 4e – Slide #6


Systems of Equations
CONTINUED
3) Determine the coordinates of the intersection point.
This ordered pair is the system’s solution. Using the
graph below, it appears that the solution is the point
(1,0). We won’t know for sure until after we check this
potential solution in the next step.



Blitzer, Intermediate Algebra, 4e – Slide #7
Systems of Equations
CONTINUED
4) Check the solution in both equations.
4x  y  4
3x  y  3
41  0  4
31  0 ? 3
?
?
40  4
4  4 true
?
3 0  3
3  3 true
Because both equations are
satisfied, (1,0) is the solution and
{(1,0)} is the solution set.
Blitzer, Intermediate Algebra, 4e – Slide #8



Substitution Method
Solving Linear Systems by Substitution
1) Solve either of the equations for one variable in terms of the other. (If
one of the equations is already in this form, you can skip this step)
2) Substitute the expression found in step 1 into the other equation. This
will result in an equation in one variable! (Hence, it’s called the
“Substitution Method”)
3) Solve the equation containing one variable. (the resultant equation
from step 2)
4) Back-Substitute the value found in step 3 into one of the original
equations. Simplify and find the value of the remaining variable.
5) Check the proposed solution in both of the system’s given equations.
Blitzer, Intermediate Algebra, 4e – Slide #9
Substitution Method
EXAMPLE
Solve by the substitution method:
4x  y  4
3x  y  3
SOLUTION
1) Solve either of the equations for one variable in terms
of the other. We’ll isolate the variable y from the first
equation.
4x  y  4
y  4 x  4
Solve for y by subtracting
4x from both sides
Blitzer, Intermediate Algebra, 4e – Slide #10
Substitution Method
CONTINUED
2) Substitute the expression from step 1 into the other
equation. -4x + 4 is the ‘expression from step 1’ and ‘the
other equation’ is 3x – y = 3. Therefore:
3x  y  3
3x   4x  4  3
Replace y with -4x+4
3) Solve the resulting equation containing one variable.
3x  4x  4  3
7x  4  3
7x  7
x 1
Distribute
Add like terms
Add 4 to both sides
Divide both sides by 7
Blitzer, Intermediate Algebra, 4e – Slide #11
Substitution Method
CONTINUED
4) Back-substitute the obtained value into one of the
original equations. We back-substitute 1 for x into one
of the original equations to find y. Let’s use the first
equation.
4x  y  4
41  y  4
4 y  4
y0
Replace x with 1
Multiply
Subtract 4 from both sides
Therefore, the potential solution is (1,0).
Blitzer, Intermediate Algebra, 4e – Slide #12
Substitution Method
CONTINUED
5) Check. Now we will show that (1,0) is a solution for
both of the original equations.
4x  y  4
3x  y  3
?


41 0  4
31  0 ? 3
?
?
40  4
3 0  3
4  4 true
3  3 true
Because both equations are
satisfied, (1,0) is the solution and
{(1,0)} is the solution set.
Blitzer, Intermediate Algebra, 4e – Slide #13
Addition (Elimination) Method
Solving Linear Systems by Addition
1) If necessary, rewrite both equations in the form Ax + By = C.
2) If necessary, multiply either equation or both equations by appropriate
nonzero numbers so that the sum of the x-coefficients or the sum of the ycoefficients is 0.
3) Add the equations in step 2. The sum should be an equation in one
variable.
4) Solve the equation in one variable (the result of step 3).
5) Back-substitute the value obtained in step 4 into either of the given
equations and solve for the other variable.
6) Check the solution in both of the original equations.
NOTE: As you now know, there is more than one method to solve a system of equations. The
reason for learning more than one method is because sometimes one method will be preferable or
easier to use over another method.
Blitzer, Intermediate Algebra, 4e – Slide #14
Addition (Elimination) Method
EXAMPLE
Solve by the addition method:
SOLUTION
4x = -2y + 4
-y = -3x + 3
1) Rewrite both equations in the form Ax + By = C. We
first arrange the system so that variable terms appear on
the left and constants appear on the right. We obtain
4x + 2y = 4
Add 2y to both sides
3x - y = 3
Add 3x to both sides
Blitzer, Intermediate Algebra, 4e – Slide #15
Addition (Elimination) Method
CONTINUED
2) If necessary, multiply either equation or both equations
by appropriate numbers so that the sum of the xcoefficients is 0. We can eliminate the y’s by multiplying
the second equation by 2. Or we can eliminate the x’s by
multiplying the first equation by -3 and the second
equation by 4. Let’s use the first method.
4x + 2y = 4
3x - y = 3
No Change
Multiply by 2
4x + 2y = 4
6x - 2y = 6
Blitzer, Intermediate Algebra, 4e – Slide #16
Addition (Elimination) Method
CONTINUED
3) Add the equations.
4x + 2y = 4
6x - 2y = 6
Add: 10x + 0y = 10
10x = 10
4) Solve the equation in one variable. Now I solve the
equation 10x = 10.
10x = 10
x=1
Blitzer, Intermediate Algebra, 4e – Slide #17
Addition (Elimination) Method
CONTINUED
5) Back-substitute and find the value of the other variable.
Now we will use one of the original equations and replace x
with 1 to determine y. I’ll use the second equation.
-y = -3x + 3
-y = -3(1) + 3
Replace x with 1
-y = -3 + 3
Multiply
-y = 0
Add
y=0
Multiply by -1
Therefore, the potential solution is (1,0).
Blitzer, Intermediate Algebra, 4e – Slide #18
Addition (Elimination) Method
CONTINUED
6) Check. I now check the potential solution (1,0) in both
original equations.
-y = -3x + 3
4x = -2y + 4
?
-(0) =? -3(1) + 3
?
0 =? -3 + 3
4(1) = -2(0) + 4
4=0+4
4=4
true
0=0
true
Because both equations are
satisfied, (1,0) is the solution and
{(1,0)} is the solution set.
Blitzer, Intermediate Algebra, 4e – Slide #19
Solving Systems of Equations
Comparing Solution Methods
Method
Advantages
You can see the solutions.
Graphing
Substitution
Addition
Disadvantages
If the solutions do not involve
integers or are too large to be
seen on the graph, it’s
impossible to tell exactly what
the solutions are.
Gives exact solutions.
Solutions cannot be seen.
Easy to use if a variable is Introduces extensive work with
on one side by itself.
fractions when no variable has a
coefficient of 1 or -1.
Gives exact solutions.
Easy to use if a variable
has a coefficient of 1 or 1.
Solutions cannot be seen.
Blitzer, Intermediate Algebra, 4e – Slide #20
Solving Systems of Equations
The Number of Solutions to a System of Two Linear
Equations
Number of Solutions
What This Means
Graphically
Exactly one ordered-pair
solution
The two lines intersect at
one point.
No solution
The two lines are
parallel.
Infinitely Many
Solutions
The two lines are
identical.
Blitzer, Intermediate Algebra, 4e – Slide #21
Graphical Examples
Solving Systems of Equations
NOTE: It is extremely helpful to understand these relationships as
well as any other relationship between an equation and it’s graph.
NOTE: To determine that a system has exactly one solution, solve
the system using one of the methods. A single solution will occur
as in the previous examples.
NOTE: To determine that a system has no solution, solve the
system using one of the methods. Eventually, you’ll get an
obviously false statement, like 3 = 4.
NOTE: To determine that a system has infinitely many solutions,
solve the system using one of the methods. Eventually, you’ll get
an obviously true statement, like -2 = -2.
Blitzer, Intermediate Algebra, 4e – Slide #22
Solving Systems of Equations
EXAMPLE
At a price of p dollars per ticket, the number of tickets to a rock
concert that can be sold is given by the demand model
N = -25p + 7800. At a price of p dollars per ticket, the number
of tickets that the concert’s promoters are willing to make
available is given by the supply model N = 5p + 6000.
(a) How many tickets can be sold and supplied for
$50 per ticket?
(b) Find the ticket price at which supply and demand
are equal. At this price, how many tickets will be
supplied and sold?
Blitzer, Intermediate Algebra, 4e – Slide #23
Solving Systems of Equations
CONTINUED
SOLUTION
(a) How many tickets can be sold and supplied for
$50 per ticket?
The number of tickets that can be sold for $50 per ticket is
found using the demand model:
N = -25(50) + 7800 = -1250 + 7800 = 6550 tickets sold.
The number of tickets that can be supplied for $50 per ticket is
found using the supply model:
N = 5(50) + 6000 = 250 + 6000 = 6250 tickets supplied.
Blitzer, Intermediate Algebra, 4e – Slide #24
Solving Systems of Equations
CONTINUED
(b) Find the ticket price at which supply and demand
are equal. At this price, how many tickets will be
supplied and sold?
1) Solve either of the equations for one variable in terms
of the other. The system of equations already has N
isolated in both equations.
N = -25p + 7800
N = 5p + 6000
Blitzer, Intermediate Algebra, 4e – Slide #25
Substitution Method
CONTINUED
2) Substitute the expression from step 1 into the other
equation. I’ll replace the N in the second equation with 25p + 7800.
-25p + 7800 = 5p + 6000
3) Solve the resulting equation containing one variable.
-25p + 7800 = 5p + 6000
7800 = 30p + 6000
1800 = 30p
60 = p
Add 25p to both sides
Subtract 6000 from both sides
Divide both sides by 30
Blitzer, Intermediate Algebra, 4e – Slide #26
Substitution Method
CONTINUED
4) Back-substitute the obtained value into one of the
original equations. We back-substitute 60 for p into
one of the original equations to find N. Let’s use the
first equation.
N = -25p + 7800
N = -25(60) + 7800
Replace p with 60
N = -1500 + 7800
Multiply
N = 6300
Add
Blitzer, Intermediate Algebra, 4e – Slide #27
Substitution Method
CONTINUED
5) Check. Now we will show that (60,6300) is a solution
for both of the original equations.
N = -25p + 7800
6300 =? -25(60) + 7800
6300 =? -1500 + 7800
6300 = 6300 true
N = 5p + 6000
6300 =? 5(60) + 6000
6300 =? 300 + 6000
6300 = 6300 true
Therefore, the solution is (60,6300). Therefore, supply and
demand will be equal when the ticket price is $60 and 6300
tickets are sold.
Blitzer, Intermediate Algebra, 4e – Slide #28