System of Linear
Equations
The whole purpose of education is
to turn mirrors into windows
~Sydney J. Harris
What is a System of Linear
Equations?
 A system of linear equations is simply two or more linear
equations using the same variables.
 We will only be dealing with systems of two equations
using two variables, x and y.
 If the system of linear equations is going to have a
solution, then the solution will be an ordered pair (x , y)
where x and y make both equations true at the same
time.
 We will be working with the graphs of linear systems and
how to find their solutions graphically.
Solve Linear Systems
Consider the following system:
y
x – y = –1
x + 2y = 5
Using the graph to the right, we
can see that any of these ordered
pairs will make the first equation
true since they lie on the line.
(1 , 2)
We can also see that any of these
points will make the second
equation true.
However, there is ONE coordinate
that makes both true at the same
time…
The solution is simply the point of intersection
x
Solution to Sample
We must always verify a proposed solution
algebraically. We propose (1,2) as a
solution, so now we plug it in to both
x=1
equations to see if it works:
y=2
x – y = -1
(1) – 2 =
- 1 = -1 
x + 2y
(1) + 2(2)
1 + 4
5
=5
=
=
= 5
Yes, (1,2) Satisfies both equations!
Solving System of Linear Equations
There are several methods of solving
systems of linear equations. Each is best
used in different situations.
Graphing Method
 Substitution Method
 Elimination (Addition) Method

Graphing Method
1.
To solve a system of linear equations by the
graphing method, there are three basic steps
to follow:
Graph the equations on the same coordinate
plane
a)
2.
Estimate where the graphs intersect.
a)
3.
Use the slope and y–intercept if needed. Be sure
to use a ruler and graph paper!
This is the solution! LABEL the solution!
Check to make sure your solution makes both
equations true.
Graphing Method
Solve the following system of equations by graphing.
y
3x + 6y = 15
–2x + 3y = –3
Step 1: Graph both equations on
the same coordinate plane
Step 2: Estimate where the
graphs intersect. LABEL the
solution!
Step 3: Check to make sure your
solution makes both equations
true.
3x + 6y = 15 -2x + 3y = -3
3(3) + 6(1) =
-2(3) + 3(1) =
9 + 6
= 15
-6 + 3 = -3
(3 , 1)
Label the
solution!
-2x
3x + 6y
3y = 15
-3
6y
3y==-3x
2x –+ 315
y = 2/3x
-1/2 x– +1 5/2
x
Types of Solutions
 If the lines cross once, there will
be one solution.
 If the lines are parallel, there will
be no solutions.
 If the lines are the same, there will
be an infinite number of solutions.
Substitution Method
Substitution method is used when it appears
easy to solve for one variable in terms of the
other. The goal is to reduce the system to
two equations of one unknown each.
Consider the following:
2x + 4y = 28
y = 3x
Substitution Method
Solve using substitution.
y = 3x
2x + 4y = 28
2x + 4(3x) = 28
2x + 12x = 28
14x = 28
x=2
y = 3x
y = 3(2)
y=6
(2,6)
Substitution Method
To solve a system of equations by substitution…
1. Solve one equation for one of the variables.
2. Substitute the value of the variable into the other
equation.
3. Simplify and solve the equation.
4. Substitute back into either equation to find the value of
the other variable.
5. Check the solution
Solve using substitution.
-3x + y = -17
y = 3x – 17
3x + 2y = 2
3x + 2(3x – 17) = 2
3x + 6x – 34 = 2
Step 1: Solve one
equation for one of the
variables
9x – 34 = 2
9x = 36
x=4
Step 2: Substitute the
value of the variable into
the other equation.
Step 3: Simplify and
solve the equation.
Step 4: Substitute back
into either equation to
find the value of the
other variable.
(4,-5)
-3x + y = -17
-3(4) + y = -17
-12 + y = -17
y = -5
Check Solution
We must always verify a
-3x + y = -17
proposed solution algebraically.
We propose (4,-5) as a solution, 3x + 2y = 2
so now we plug it in to both
equations to see if it works:
-3x + y
-3(4) + (-5)
-12 – 5
-17
= -17
=
=
= -17 
3x + 2y
3(4) + 2(-5)
12 + -10
2
=2
=
=
=2 
Yes, (4,-5) Satisfies both equations!
Elimination Method
Elimination method is used when it appears
easy to eliminate one variable from the system
by adding the two equations together
The elimination method makes use of the
addition principle of equality
if a = b, then a + c = b + c
Elimination Method
Solve using the elimination method.
x–y=7
x+y=3
x+y=3
5+y=3
y = -2
2x + 0y = 10
2x = 10
x=5
(5,-2)
Elimination Method
To solve a system of equations by elimination…
1. Put equations in standard form (Ax + By = C)
2. Determine which variable to eliminate.
3. Add or subtract the equations and solve for the variable.
4. Substitute back into either equation to find the value of
the other variable.
5. Check the solution
Solve using the elimination method.
2x + 3y = 11
-2x + 9y = 1
0x + 12y = 12
12y = 12
y=1
2x + 3y = 11
2x + 3(1) = 11
2x + 3 = 11
2x = 8
x=4
(4,1)
Elimination Method
Solve using the elimination method.
2x + 3y = 11
-2x + 9y = 1
0x +12y = 12
12y = 12
y=1
Step 1: Put in standard form
2x + 3y = 11
2x + 3(1) = 11
2x + 3 = 11
2x = 8
x=4
Done
Step 2: Determine which variable to eliminate Variable x
Step 3: Add equations and solve
Step 4: Substitute and solve for other variable
(4,1)
Check Solution
We must always verify a
2x + 3y = 11
proposed solution algebraically.
-2x
+
9y
=
1
We propose (4,1) as a solution,
so now we plug it in to both
equations to see if it works:
2x + 3y
2(4) + 3(1)
8 + 3
11
= 11
=
=
= 11 
-2x + 9y
-2(4) + 9(1)
-8 + 9
1
=1
=
=
=1 
Yes, (4,1) Satisfies both equations!
Elimination Method
Sometimes you may need to utilize the
multiplication property of equality
if a = b, then ac = bc
to help eliminate a variable
Solve using the addition method.
3x – y = 8
x + 2y = 5
(2)(3x – y)=(8 )(2)
x + 2y = 5
6x – 2y = 16
x + 2y = 5
7x
= 21
x=3
write in standard form
multiply as needed
add the equations
substitute
x + 2y = 5
(3) + 2y = 5
2y = 2
y=1
(3,1)
Solve using the addition method.
3x + 5y = 12
4x – 3y = -13
(3)(3x + 5y )= ( 12)(3)
(5)(4x – 3y )= (-13)(5)
9x + 15y = 36
20x – 15y = -65
29x
= -29
x = -1
write in standard form
multiply as needed
(eliminate x variable)
add the equations
substitute
3x + 5y = 12
3(-1) + 5y = 12
5y = 15
y=3
(-1,3)
Parallel Lines (no solution)
y = -2x – 3
y = -2x + 5
(-1)( 2x + y) = (-3)(-1)
2x + y = 5
-2x – y = 3
2x + y = 5
0 +0=8
0=8
write in standard form
multiply as needed
add the equations
This is a contradiction
since 0 does not
equal to 8
No solution
Same Lines (infinite solutions)
6x + 2y = 4
y = -3x + 2
6x + 2y = 4
(-2)( 3x + y) = (2 )(-2)
6x + 2y = 4
-6x – 2y = -4
0 + 0=0
0=0
write in standard form
multiply as needed
add the equations
This is a always a true
statement regardless
of values for x and y
Infinite solutions
The sum of a number and twice another
number is 13. The first number is 4 larger than
the second number. What are the numbers?
Let x = the first number
Let y = the second number
x + 2y = 13
x=y+4
Use substitution
y + 4 + 2y = 13
3y + 4 = 13
3y = 9
y=3
x=y+4
x=3+4
x=7
The admission fee at a small fair is $1.00 for children
and $4.00 for adults. On a certain day, 1,000 people
entered the fair and $2,200 is collected. How many
children and how many adults attended?
Let x = number of adults
Let y = number of children
Use elimination
x + y = 1000
400 + y = 1000
y = 600 children
(-1)( x + y )= (1000)(-1)
4x + y = 2200
-x – y = -1000
4x + y = 2200
3x
= 1200
x = 400 adults