Systems of Linear Equations
A system of linear equations consists of two or more
linear equations. We will focus on only two equations at
a time.
The solution of a system of linear equations in two
variables is any ordered pair that solves both of the
linear equations. The solution to the system is the point
that satisfies ALL of the equations. This point will be an
ordered pair.
When graphing, you will encounter three possibilities.
IDENTIFYING THE NUMBER OF SOLUTIONS
NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
y
y
y
x
x
x
Lines intersect
Lines are parallel
Lines coincide
one solution
no solution
infinitely many solutions
Parallel Lines
 These lines never
intersect!
 Since the lines never
cross, there is
NO SOLUTION!
 Parallel lines have the
same slope with different
y-intercepts.
2
=2
1
y-intercept = 2
y-intercept = -1
Slope =
Same Lines
 These lines are the same!
 Since the lines are on top
of each other, there are
INFINITELY MANY
SOLUTIONS!
 Coinciding lines have the
same slope and
y-intercepts.
2
=2
1
y-intercept = -1
Slope =
What is the solution of the system
graphed below?
1. (2, -2)
2. (-2, 2)
3. No solution
4. Infinitely many solutions
Name the Solution
Solving a system of equations by graphing.
Let's summarize! There are 3 steps to solving a system
using a graph.
Step 1: Graph both equations.
Graph using slope and y – intercept
or x- and y-intercepts. Be sure to use
a ruler and graph paper!
Step 2: Do the graphs intersect?
This is the solution! LABEL the
solution!
Step 3: Check your solution.
Substitute the x and y values into
both equations to verify the point is a
solution to both equations.
Solution of a System
Example
Determine whether the given point is a solution of the
following system.
point: (– 3, 1)
system: x – y = – 4 and 2x + 10y = 4
•Plug the values into the equations.
First equation: – 3 – 1 = – 4 true
Second equation: 2(– 3) + 10(1) = – 6 + 10 = 4
true
•Since the point (– 3, 1) produces a true statement in both
equations, it is a solution.
1) Find the solution to the following
system:
2x + y = 4
x-y=2
Graph both equations. I will graph using
x- and y-intercepts (plug in zeros).
2x + y = 4
(0, 4) and (2, 0)
x–y=2
(0, -2) and (2, 0)
Graph the ordered pairs.
Graph the equations.
2x + y = 4
(0, 4) and (2, 0)
x-y=2
(0, -2) and (2, 0)
Where do the lines intersect?
(2, 0)
Check your answer!
To check your answer, plug
the point back into both
equations.
2x + y = 4
2(2) + (0) = 4
x-y=2
(2) – (0) = 2
Nice job…let’s try another!
2) Find the solution to the following
system:
y = 2x – 3
-2x + y = 1
Graph both equations. Put both equations in slopeintercept or standard form. I’ll do slope-intercept form
on this one!
y = 2x – 3
y = 2x + 1
Graph using slope and y-intercept
Graph the equations.
y = 2x – 3
m = 2 and b = -3
y = 2x + 1
m = 2 and b = 1
Where do the lines intersect?
No solution!
Notice that the slopes are the same with different
y-intercepts. If you recognize this early, you don’t
have to graph them!
Practice – Solving by Graphing



(1,2)
y – x = 1  (0,1) and (-1,0)
y + x = 3  (0,3) and (3,0)
Solution is probably (1,2) …
Check it:
2 – 1 = 1 true
2 + 1 = 3 true
therefore, (1,2) is the solution
Practice – Solving by
Graphing
Inconsistent: no solutions



y = -3x + 5  (0,5) and (3,-4)
y = -3x – 2  (0,-2) and (-2,4)
They look parallel: No solution
Check it:
m1 = m2 = -3
Slopes are equal
therefore it’s an inconsistent
system
Consistent: infinite sol’s



3y – 2x = 6  (0,2) and (-3,0)
-12y + 8x = -24  (0,2) and (-3,0)
(1,2)
Looks like a dependant system …
Check it:
divide all terms in the 2nd equation by -4
and it becomes identical to the 1st
equation
therefore, consistent, dependant system
Graph the system of equations. Determine whether the system has
one solution, no solution, or infinitely many solutions. If the system
has one solution, determine the solution.
1.
x  3y  3
3x  9 y  9
3
2. y  x  4
5
5 y  3x
3.
x y3
2x  y  6
The Substitution Method
Solving a System of Linear Equations by the
Substitution Method
1)
2)
3)
4)
5)
Solve one of the equations for a variable.
Substitute the expression from step 1 into the other
equation.
Solve the new equation.
Substitute the value found in step 3 into either
equation containing both variables.
Check the proposed solution in the original
equations.
The Substitution Method
Solve the following system using the substitution method.
3x – y = 6 and – 4x + 2y = –8
Solving the first equation for y,
3x – y = 6
–y = –3x + 6
y = 3x – 6
(subtract 3x from both sides)
(multiply both sides by – 1)
Substitute this value for y in the second equation.
–4x + 2y = –8
–4x + 2(3x – 6) = –8
–4x + 6x – 12 = –8
2x – 12 = –8
2x = 4
x=2
(replace y with result from first equation)
(use the distributive property)
(simplify the left side)
(add 12 to both sides)
(divide both sides by 2)
Continued.
The Substitution Method
Substitute x = 2 into the first equation solved for y.
y = 3x – 6 = 3(2) – 6 = 6 – 6 = 0
Our computations have produced the point (2, 0).
Check the point in the original equations.
First equation,
3x – y = 6
3(2) – 0 = 6
true
Second equation,
–4x + 2y = –8
–4(2) + 2(0) = –8
true
The solution of the system is (2, 0).
The Elimination Method
Solving a System of Linear Equations by the Addition
or Elimination Method
1)
2)
3)
4)
5)
6)
Rewrite each equation in standard form, eliminating
fraction coefficients.
If necessary, multiply one or both equations by a number
so that the coefficients of a chosen variable are opposites.
Add the equations.
Find the value of one variable by solving equation from
step 3.
Find the value of the second variable by substituting the
value found in step 4 into either original equation.
Check the proposed solution in the original equations.
The Elimination Method
Solve the following system of equations using the elimination
method.
6x – 3y = –3 and 4x + 5y = –9
Multiply both sides of the first equation by 5 and the second
equation by 3.
First equation,
5(6x – 3y) = 5(–3)
30x – 15y = –15
(use the distributive property)
Second equation,
3(4x + 5y) = 3(–9)
12x + 15y = –27
(use the distributive property)
Continued.
The Elimination Method
Combine the two resulting equations (eliminating
the variable y).
30x – 15y = –15
12x + 15y = –27
42x
= –42
x = –1
(divide both sides by 42)
Continued.
The Elimination Method
Substitute the value for x into one of the original
equations.
6x – 3y = –3
6(–1) – 3y = –3
(replace the x value in the first equation)
–6 – 3y = –3
(simplify the left side)
–3y = –3 + 6 = 3
(add 6 to both sides and simplify)
y = –1
(divide both sides by –3)
Our computations have produced the point (–1, –1).
Continued.
The Elimination Method
Check the point in the original equations.
First equation,
6x – 3y = –3
6(–1) – 3(–1) = –3
Second equation,
4x + 5y = –9
4(–1) + 5(–1) = –9
true
true
The solution of the system is (–1, –1).