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Solving Systems of Equations
By the Graphing Method
A linear equation in two variables has an infinite number of
solutions. The set of all solutions to a linear equation forms a line
in a rectangular coordinate system. Two or more linear equations
from a system of linear equations.
Infinite number of solutions
A solution to a system of linear
equations is an ordered pair that is a
solution to each individual linear
equation.
Determining Solutions to a
System of Linear Equations
x+y=4
a). (3, 1) It is important to test an
ordered pair in both the
-2x + y = -5
equations to determine if
Solution:
the ordered pair is a
solution
a. Substitute the ordered pair (3, 1)
x+y=4
-2x + y = -5
(3) + (1) = 4
-2(3) + (1) = -5
TRUE
TRUE
Because the ordered pair (3, 1) is a solution to each equation, it
is a solution to the system of equations
Determining Solutions to a
System of Linear Equations
x+y=4
-2x + y = -5
b). (0, 4)
Solution:
a. Substitute the ordered pair (0, 4)
x+y=4
-2x + y = -5
(0) + (4) = 4
-2(0) + (4) = -5
TRUE
False
Because the ordered pair (0, 4) is not a solution to each
equation, it is not a solution to the system of equations
A solution to a system of two linear equations may
be interpreted graphically as a point of intersection
between the two lines. Using slope-intercept form to
graph the lines from Example 1, we have
x+y=4
-2x + y = -5
y = -x + 4
y = 2x - 5
Notice that the lines
intersect at (3, 1)
Two Equations Containing Two Variables
The first two cases are called consistent since there are
solutions. The last case is called inconsistent.
The solution will be one of three cases:
1. Exactly one solution, an ordered pair (x, y)
2. A dependent system with infinitely many solutions
3. No solution
independent
With two equations and two variables,
the graphs are lines and the solution (if
there is one) is where the lines
intersect. Let’s look at different
possibilities.
dependent
consistent
Case 1: The lines
intersect at a point
(x, y),
the solution.
inconsistent
consistent
Case 2: The lines
coincide and there
are infinitely many
solutions (all points
on the line).
Case 3: The lines
are parallel so there
is no solution.
Example 2
y = 2x
y=2
The lines intersect at (1, 2)
Solution:
The equation y = 2x is written in slope-intercept form as y = 2x + 0.
The line passes through the origin, with a slope of 2
The line y = 2 is a horizontal line and has a slope of 0.
EXAMPLE 3
x – 2y = -2
-3x + 2y = 6
Equation 1.
x – 2y = -2
-2y = -x -2
-2y = -x – 2
-2
-2 -2
y = 1x +1
2
Solution:
To graph each equation, write the equation in
slope-intercept form: y = mx + b
Equation 2.
-3x + 2y = 6
2y = 3x + 6
2y = 3x +6
2
2
2
y = 3x +3
2
The point (-2, 0) appears
to be the point of
intersection. This can be
confirmed by substituting
x = -2 and y = 0 into both
equations.
x – 2y = -2
-3x + 2y = 6
(-2) – 2(0) = -2
True
-3(-2) + 2(0) = 6
True
EXAMPLE 4
-x + 3y = -6
6y = 2x + 6
Equation 1.
-x + 3y = -6
3y = x -6
3y = x – 6
3
3 3
y=1x-2
3
Solution:
To graph each equation, write the equation in
slope-intercept form: y = mx + b
Equation 2.
6y = 2x + 6
6y = 2x +6
6
6 6
y = 1x +1
3
Because the lines have the same slope but different
y-intercepts, they are parallel. Two parallel line do not
intersect, which implies that the system has no solution.
The system is inconsistent.
EXAMPLE 5
x +4y = 8
y = -1 x + 2
4
Equation 1.
x + 4y = 8
4y = -x + 8
4y = -x + 8
4
4 4
y = -1x +2
4
Solution:
To graph each equation, write the equation in
slope-intercept form: y = mx + b
Equation 2.
y = -1x + 2
4
Notice that the slopeintercept forms of the
two lines are identical.
Therefore the
equations represent the
same line. The system
is dependent, and the
solution to the system
of equations is the set
of all points
Solving System of Equations by
the Substitution Method
Solve the system using the substitution method
x+y=4
-2x + y = -5
Solution:
x+y=4
x=4-y
-2 x + y = -5
-2( 4 – y ) + y = -5
This equation now contains only one
variable.
Solve the resulting equation
-8 +2y +y = -5
-8 + 3y = -5
3y = -5 + 8
y=1
To find x, substitute y = 1
back into the first equation
x=4-y
x = 4 – (1)
x=3
The ordered pair (3, 1) can be checked in the
original equations to verify the answer
x+y=4
-2x + y = -5
(3) + (1) = 4
-2(3) + (1) = -5
The solution is (3, 1)
True
True
Solving a System of Linear Equations
Using the Substitution Method
1. Isolate one of the variables from one equation
2. Substitute the quantity found in Step 1 into the
other equation.
3. Solve the resulting equation.
4. Substitute the value found in Step 3 back into
the equation in Step 1 to find the value of the
remaining variable.
5. Check the solution in both original equations
and write the answer as an ordered pair.
If we have two equations and variables and we want to
solve them, graphing would not always be very accurate so
we will solve algebraically for exact solutions. We'll look at
two methods. The first is solving by substitution.
The idea is to solve for one of the variables in one of the equations
and substitute it in for that variable in the other equation.
3x  5 y  17
Let's solve for y in the second equation. You
can pick either variable and either equation, but
go for the easiest one (getting the y alone in the
second equation will not involve fractions).
2 x  y  6
yy22 x1 66  4 Substitute this for y in the first equation.
Easiest here since we already solved for y.
3x  5  2 x  6  17
13x  13 x  1
Now we only have the x variable and we
solve for it.
Substitute this for x in one of the equations to find y.
3x  5 y  17
2 x  y  6
 1, 4
So our solution is
 1, 4
This means that the two lines intersect at
this point. Let's look at the graph.
We can check this by
subbing in these values
for x and y. They should
make each equation true.
3  1  5  4   17
2  1   4   6
3  20  17
2  4  6
Yes! Both equations are
satisfied.
Solving an Inconsistent System
Using Substitution
The idea is to solve for one of the variables in one of the equations
and substitute it in for that variable in the other equation.
2x  3y  6
2
y  x4
3
The second equation is in the slope-intercept
form.
Substitute this for y in the first equation.
 2

2x  3  x  4   6
 3

2x  2x 12  6
12  6
Contradiction There is no
solution
2x  3y  6
2
y  x4
3
There is no solution
This means that the two lines do not
intersect at any point. Let's look at the
graph.
We can check this by
writing each equation in
slope-intercept form and
graphing the lines.
2x  3y  6
3 y  2 x  6
2
y   x6
3
2
y  x4
3
Yes! Both equations have
the same slope but different
y-intercepts
Solving a Dependent System
Using Substitution
½x - ¼y = 1
6x – 3y = 12
½x - ¼y = 1
Clear the fractions
4(½x) -4( ¼y) =4(1)
2x – y = 4
Now the system becomes:
2x – y = 4
The y-variable in the first equation is the easiest
6x – 3y = 12
to isolate because its coefficient is -1
The idea is to solve for one of the variables in one of the equations
and substitute it in for that variable in the other equation.
2x  y  4
6 x  3 y  12
y  2x  4
Let's solve for y in the first equation. You can
pick either variable and either equation, but go
for the easiest one (getting the y alone in the
first equation will not involve fractions).
Substitute this for y in the second equation.
Now we only have the x variable and we
solve for it.
6 x  3  2 x  4  12
6 x  6 x  12  12
12  12
Because the equation produces
an identity, all values of x make
this equation true.
2x  y  4
6 x  3 y  12
There is an infinite number of solutions
This means that the two lines intersect at
every point. Let's look at the graph.
We can check this by
writing each equation in
slope-intercept form and
graphing the lines.
2x  y  4
6 x  3 y  12
 y  2 x  4 3 y  6 x  12
y  2x  4
y  2x  4
Yes! Both equations have the
same slope and y-intercepts.
Now let's look at the second method, called the method of
elimination.
The idea is to multiply one or both equations by a constant (or
constants) so that when you add the two equations together, one of
the variables is eliminated.
2 x  3 y  1
3
4x  y  3 3
12 x  3 y  9
2 x  3 y  1
14 x  8
4
x
7
Let's multiply the bottom equation by 3. This
way we can eliminate y's. (we could instead
have multiplied the top equation by -2 and
eliminated the x's)
Add first equation to this.
The y's are eliminated.
4
4   y  3
7
5
y
7
Substitute this for x in one of the equations to find y.
Solving Systems of Equations by
the Addition Method
Notice that the coefficients of the y-variables are
opposites:
Coefficient is 1.
x + y = -2
x – y = -6
Coefficient is -1
Because the coefficients of the y-variables are opposites, we
can add the two equations to eliminate the y-variable.
x + y = -2
x – y = -6
2x = -8
After adding the equations, we have
one equation and one variable.
Solve the resulting equation.
2x = -8
2
2
x = -4
To find the value of y, substitute x = -4 into
either of the original equations
x + y = -2 First equation
(-4) + y = -2
The solution
y = -2 + 4
is (-4, 2)
y=2
Solving a System of Equations
by the Addition Method
• Write both equations in standard form: Ax + By = C
• Clear fractions or decimals
• Multiply one or both equations by nonzero constants to
create opposite coefficients for one of the variables.
• Add the equations from Step 3 to eliminate one
variable.
• Solve for the remaining variable.
• Substitute the known value from Step 5 into one of the
original equations to solve for the other variable.
• Check the solution in both equations
So we can arrive at the same answer with either method, but
which method should you use?
It depends on the problem. If substitution would involve
messy fractions, it is generally easier to use the addition
method. However, if one variable is already or easily
solved for, substitution is generally quicker.
With either method, we may end up with a surprise. Let's see
what this means.
3 x  6 y  15
3 x  2 y  5 3
3 x  6 y  15
3 x  6 y  15
00
Let's multiply the second equation by 3 and
add to the first equation to eliminate the x's.
Everything ended up eliminated. This tells us
the equations are dependent. This is Case 2
where the lines coincide and all points on the
line are solutions.
3 x  6 y  15
x  2 y  5
x  2y 5
Let's solve the second equation for x. (Solving for x
in either equation will give the same result)
Now to get a solution, you chose any real
number for y and x depends on that choice.
y
If y is 0, x is -5 so the
point (-5, 0) is a solution
to both equations.








If y is 2, x is -1 so the
point (-1, 2) is a solution
to both equations.

Any point on
this line is a
solution
        










x






So we list the solution as:
x  2 y  5 where x is any real number
Example 6
Solve the system using the addition method
3(x – 10) = 7y + 11
-2(x – y) = 2x - 18
Step 1. Write the equations in standard form.
Clear the parentheses
3x –-10)
3(x
30 =7y
= 7y++1111
-2(x+–2y
-2x
y) == 2x
2x --18
18
Write as Ax + By = C
3x – 7y = 41
-4x + 2y = -18
Step 2. There are no fractions or decimals. No need to clear the
fractions or decimals in this equation.
Notice that neither the coefficients of x nor the coefficients
of y are opposites. However, it is possible to change the
coefficients of x to 12 and -12 (notice that 12 is the LCM
of 3 and 4). This is accomplished by multiplying the first
equation by 4 and the second equation by 3
Step 3: Create opposite
coefficients of x
Step 4: Add the
equations.
Step 5: Solve the
resulting equation.
12x – 28yMultiply
= 164 by 4
-12x
+28y
6y==41-54
12x
3x–-7y
164
110
-12x
-4x++-22y
6y
2y===-54
-18
Multiply by 3
-22y = 110
-22 -22
y = -5
3 x – 7 -5
y = 41 First equation
3x + 35 = 41
3x + 35 -35 =41 -35
3x = 6
3 3
x=2
3(x– –10)
10)
7y ++11
3(2
= =7(-5)
11
-2(x
– y) == 2(2)
2x - 18
2[2
– (-5)]
- 18
Step 6: Substitute y = -5
-5
into one of the equations.
The solution is (2, -5)
Step 7: Check the solution
in the original equations.
-24 = -24
-14 = -14
True
True
Summary of Methods for Solving Linear
Equations in Two Variables
If no method of solving a system of linear equations is specified,
you may use the method of your choice. However, it is
recommended you use the following guidelines:
1. If one of the equations is written with a variable isolated, the
substitution method is a good choice.
2x + 5y = 2
x=y-6
2. If both equations are written in standard form, Ax + By = C,
where none of the variables has coefficients of 1 or -1, then
the addition method is a good choice.
4x + 5y = 12
5x + 3y = 15
3. If both equations are written in standard form,
Ax + By = C, and at least one variable has a coefficient of 1
or -1, then either the substitution method or the addition
method is a good choice.