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Solving Systems
of Equations in
Two Variables
Section 2-1
Before finishing this section
you should be able to:
• Solve systems of equations
graphically
• Solve systems of equations
algebraically
A system of equations is a set
of two or more equations.
Solving a system of equations
means finding the ordered pair
that makes both equations true.
This is called the solution to
the system of equations. If a
system of equations were
graphed, the solution would be
the point where the lines
intersect.
Solve the system of equations by
graphing.
2x + 5y = 9
x-y=1
First rewrite each equation of the system
in slope-intercept form by solving for y.
2x + 5y = 9 
x-y=1

y=
2
9
 x
5
5
y=x-1
Since the two lines have different
slopes, the graphs of the
equations are intersecting lines.
The solution to the system is
(2, 1)
Classifications of systems
of equations:
A consistent system of equations has at least one
solution. If there is exactly one solution, the system is
independent. The previous example consistent (at
least one solution) and independent (exactly one
solution).
Recall, the graphs of two equations may be the same
line. In this case, there are infinitely many solutions
and the system is dependent (also consistent
because there is at least one solution).
Lines may not always intersect. They do not intersect if
the lines are parallel. Therefore, there is no solution
and the system is inconsistent. The chart on the top
of pg. 68 in your textbook summarizes the
characteristics of these types of systems.
A system of equations may be solved
algebraically.
Two common methods are the elimination
method and the substitution method.
The elimination method is when the equations
are stacked and added together. One or both
equations may have to be manipulated by
multiplying the entire equation by a constant.
The goal when adding the equations is that one
of the variables is eliminated.
Solving by using Elimination
Use the elimination method to solve the system of
equations.
3x - 2y = 18
4x + 3y = -10
OneOne
way
to to
solve
way
solvethis
thissystem
systemisistoto multiply
multiply both
both sides
sides ofofthe first equation by 3, m
the both
first sides
equation
3, multiply
both
thethe
second
of thebysecond
equation
bysides
2, andofadd
two equations to eliminate y. T
equation
by 2, and
add the two equations to eliminate y.
the resulting
equation.
Then solve the resulting equation.
3(3x - 2y) = 3(18)
2(4x + 3y) = 2(-10)
x


9x - 6y = 54
8x + 6y = -20
17x
= 34
=2
Now substitute 2 for x in either of the original equations.
3x - 2y = 18
3(2) - 2y = 18
x=2
-2y = 12
y
= -6
The solution is (2, -6). Check it by
substituting into 4x + 3y = -10. If the
coordinates make both equations true,
then the solution is correct.
In the SUBSTITUTION METHOD, first we
must solve one of the equations for either
x or y. (Hint: it will be easier to solve for
the variable that has a coefficient of 1)
Next, substitute what x equals into the
other equation. Solve this equation for y.
Substitute the solution for y into either of
the two equations and solve for x.
Solving by using Substitution
Use the substitution method to solve the
system of equations.
4x - 3y= 11
x+y = 8
You can solve the second equation for either y or x. If you
solve for x, the result is x = 8 - y.
Then substitute 8 - y for x in the first equation.
4x - 3y = 11
4(8 - y) - 3y = 11
x=8-y
-7y = -21
y
=3
Now substitute 3 for y in either of the original equations, and
solve for x.
x + y= 8
x+3 =8
x
y=3
=5
The solution is (5, 3)
Real-World Example
CONSUMER CHOICES Jeremy is considering
two different cell phone plans. The first plan
has a $25 monthly fee plus $0.25 per minute
used. The second plan offers a $10 monthly
fee with a $0.40 charge per minute used.
a. What is the break-even point in the two cell
phone plans that Jeremy is considering?
b. If Jeremy expects to use the phone for no
more than 75 minutes each month, which
plan should he choose?
a. First, write an equation to represent
the amount he will pay with each
plan. Let C represent the total
monthly cost and m represent the
number of minutes used.
Plan 1 ($25 monthly charge plus $0.25
per minute): C = 0.25m + 25
Plan 2 ($10 monthly charge plus $0.40
per minute): C = 0.40m + 10
Now, solve the system of equations.
Since both equations contain C, we
can substitute the value of C from
one equation into the other.
C
0.40m + 10
0.15m
m
= 0.25m + 25
= 0.25m + 25
= 15
= 100
C = 0.40m + 10
The break-even point occurs at a monthly usage of 100
minutes.
b.
The graph of the equations shows that for
monthly usage under 100 minutes, Plan 2 is less
expensive. So, Jeremy should probably choose Plan 2.
Helpful Websites
Solving systems by graphing:
• http://regentsprep.org/Regents/math/syslin/GrSys.htm
• http://regentsprep.org/Regents/math/syslin/PracGr.htm
• http://regentsprep.org/Regents/math/syslin/TSys.htm
Solving systems by algebra:
• http://regentsprep.org/Regents/math/syslin/AlgSys.htm
• http://regentsprep.org/Regents/math/syslin/AlgSysAdd.htm
•http://www.studyworksonline.com/cda/content/article/0,,NA
V14-105_SAR1864,00.shtml
• http://as.clayton.edu/garrison/Math%200099/Solving_Systems_of_E
quations.htm
2-1 Self-Check Quiz:
• http://www.glencoe.com/sec/math/studytools/cgibin/msgQuiz.php4?isbn=0-07-8608619&chapter=2&lesson=1&quizType=1&headerFile=4&state=