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4.1
Solving Systems of
Linear Equations in
Two Variables
Systems of Linear Equations
A system of equations consists of two or more
equations.
The solution of a system of two equations in two
variables is an ordered pair (x, y) that makes both
equations true.
Example
Determine whether (–3, 1) is a solution of the system.
x–y=–4
2x + 10y = 4
Replace x with –3 and y with 1 in both 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 of the system.
Example
Determine whether (4, 2) is a solution of the system.
2x – 5y = – 2
3x + 4y = 4
Replace x with 4 and y with 2 in both equations.
First equation: 2(4) – 5(2) = 8 – 10 = – 2
True
Second equation: 3(4) + 4(2) = 12 + 8 = 20 ≠ 4 False
Since the point (4, 2) produces a true statement in only one
equation, it is NOT a solution.
Solving Systems of Equations by
Graphing
Since a solution of a system of equations is a
solution common to both equations, it is also a
point common to the graphs of both equations.
To find the solution of a system of two linear
equations, we graph the equations and see where
the lines intersect.
Example
Solve the system of
equations by graphing.
2x – y = 6
x + 3y = 10
First, graph 2x – y = 6.
Second, graph x + 3y = 10.
The lines APPEAR to
intersect at (4, 2).
Y
10 Y Y
109
10 Y
10
989
9
878
8
767
7
656
6
545
5
434
4
323
3
212
2
X
1 1
X X
10
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-1
1 2 3 4 5 6 7 8 9 10 X
0 1 12 23 34 45 56 67 78 89 10
-10-10
-9 -9
-8 -8
-7 -7
-6 -6
-5 -5
-4 -4
-3 -3
-2 -2
-1-10-1-1
10
0 1 2 3 4 5 6 7 8 9910
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-2
-1
-2-3-2
-2
-3-4-3
-3
-4-5-4
-4
-5-6-5
-5
-6-7-6
-6
-7-8-7
-7
-8-9-8
-8
-9
-9
-10
-9
-10-10
-10
continued
Example (cont)
Although the solution to the system of equations
appears to be (4, 2), you still need to check the
answer by substituting x = 4 and y = 2 into the two
equations.
First equation:
2(4) – 2 = 8 – 2 = 6
True
Second equation:
4 + 3(2) = 4 + 6 = 10 True
The point (4, 2) checks, so it is the solution of the
system.
Helpful Hint
Neatly drawn graphs can help when “guessing”
the solution of a system of linear equations by
graphing.
Example
Solve the system of
equations by graphing.
–x + 3y = 6
3x – 9y = 9
First, graph – x + 3y = 6.
Second, graph 3x – 9y = 9.
The lines APPEAR to
be parallel.
Y
10 Y
10 Y
10
9
9
9
8
8
8
7
7
7
6
6
6
5
5
5
4
4
4
3
3
3
2
2
2
1
X
1
X
1
X
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10
1
2
3
4
5
6
7
8
9
10
0
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10 1 2 3 4 5 6 7 8 9 10
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-1
1 2 3 4 5 6 7 8 9 10
-2
-2
-2
-3
-3
-3
-4
-4
-4
-5
-5
-5
-6
-6
-6
-7
-7
-7
-8
-8
-8
-9
-9
-9
-10
-10
-10
continued
Example (cont)
Although the lines appear to be parallel, we need to check their slopes.
–x + 3y = 6
3y = x + 6
1
3
y= x+2
3x – 9y = 9
–9y = –3x + 9
1
y= x–1
3
First equation
Add x to both sides.
Divide both sides by 3.
Second equation
Subtract 3x from both sides.
Divide both sides by –9.
1
Both lines have a slope of , so they are parallel and do
3
not intersect. Hence, there is no solution to the system.
Example
Solve the system of
equations by graphing.
x = 3y – 1
2x – 6y = –2
First, graph x = 3y – 1.
Second, graph 2x – 6y = –2.
The lines APPEAR to
be identical.
YY
Y
10
10
10
99
9
88
8
77
7
66
6
55
5
44
4
33
3
22
2
11
XX
1
X
00 11 22 33 44 55 66 77 88 9910
-10
-10-9-9-8-8-7-7-6-6-5-5-4-4-3-3-2-2-1-1
-1-1
10
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10 1 2 3 4 5 6 7 8 9 10
-2-2
-2
-3-3
-3
-4-4
-4
-5-5
-5
-6-6
-6
-7-7
-7
-8-8
-8
-9-9
-9
-10
-10
-10
continued
Example (cont)
Although the lines appear to be identical, we need to check that their
slopes and y-intercepts are the same.
x = 3y – 1
First equation
3y = x + 1
Add 1 to both sides.
y = 1x + 1
3
Divide both sides by 3.
3
2x – 6y = – 2
–6y = – 2x – 2
y
1
=
3
x+
1
3
Second equation
Subtract 2x from both sides.
Divide both sides by -6.
Any ordered pair that is a solution of one equation is a solution
of the other. This means that the system has an infinite
number of solutions.
Identifying Special Systems of
Linear Equations
There are three possible outcomes when graphing
two linear equations in a plane.
One point of intersection—one solution
Parallel lines—no solution
Coincident lines—infinite number of solutions
If there is at least one solution, the system is
considered to be consistent.
If the system defines distinct lines, the equations are
independent.
Possible Solutions of Linear
Equations
Graph
Number of Solutions Type of System
(3, 5)
If the lines intersect, the Consistent
system of equations has The equations are
one solution given by the independent.
point of intersection.
Two lines intersect at one point.
Parallel lines
Lines coincide
If the lines are parallel,
then the system of
equations has no
solution because the
lines never intersect.
Inconsistent
The equations are
independent.
If the lines lie on top of
each other, then the
system has infinitely
many solutions. The
solution set is the set of
all points on the line.
Consistent
The equations are dependent.
The Substitution Method
Another method that can be used to solve systems of
equations is called the substitution method.
To use the substitution method, we first need an equation
solved for one of its variables. Then substitute that new
expression for the variable into the other equation and solve
for the other variable.
Example
Solve the system using the substitution method.
3x – y = 6
– 4x + 2y = –8
The first equation can be easily solved for y.
3x – y = 6
–y = –3x + 6
y = 3x – 6
Subtract 3x from both sides.
Multiply both sides by –1.
Now we 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.
Add 12 to both sides.
Divide both sides by 2.
continued
Example (cont)
Substitute x = 2 into the first equation, which has already
been solved for y.
y = 3x – 6 = 3(2) – 6 = 6 – 6 = 0
The solution of the system is the ordered pair (2, 0).
Check the point (2, 0) 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 Substitution Method
Solving a System of Two Equations Using the
Substitution Method
Step 1: Solve one of the equations for one of its variables.
Step 2: Substitute the expression for the variable found in
Step 1 into the other equation.
Step 3: Find the value of one variable by solving the
equation from Step 2.
Step 4: Find the value of the other variable by substituting
the value found in Step 3 into the equation from
Step 1.
Step 5: Check the ordered pair solution in both original
equations.
Example
Solve the system:
y = 2x – 5
8x – 4y = 20
Since the first equation is already solved for y, substitute
this value into the second equation.
8x – 4y = 20
8x – 4(2x – 5) = 20
Replace y with result from first equation.
8x – 8x + 20 = 20
Use distributive property.
20 = 20
Simplify.
continued
Example (cont)
A true statement such as 20 = 20 indicates that the two
linear equations actually represent the same line.
There are an infinite number of solutions for this
system. Any solution of one equation would
automatically be a solution of the other equation.
This represents a consistent system and the linear
equations are dependent equations.
Example
Solve the following system of equations:
3x – y = 4
6x – 2y = 4
Solve the first equation for y.
3x – y = 4
–y = –3x + 4
Subtract 3x from both sides.
y = 3x – 4
Multiply both sides by –1.
Substitute this value for y into the second equation.
6x – 2y = 4
6x – 2(3x – 4) = 4
Replace y with the result from the first equation.
6x – 6x + 8 = 4
Use the distributive property.
8=4
Simplify.
continued
Example (cont)
The false statement 8 = 4 indicates that this system has
no solution. The graph of the linear equations in the
system would be a pair of parallel lines.
This represents an inconsistent system, even though
the linear equations are independent.
Solving a System Using Elimination
Another method that can be used to solve systems
of equations is called the addition or elimination
method.
You multiply both equations by numbers that will
allow you to combine the two equations and
eliminate one of the variables.
Example
Solve the following system
x+y=7
x–y=9
x+y= 7
x–y= 9
2x =
16
x=8
Add the equations to eliminate y.
Divide both sides by 2.
continued
Example (cont)
Substitute 8 for x into one of the original equations and solve
for y.
x+y=7
8+y=7
y = –1
Replace the x value with 8 in the first equation.
Add 8 to both sides of the equation.
Our computations have produced the point (8, –1).
continued
Example (cont)
Check the point in the original equations.
x+y= 7
8 + (–1) = 7
True
x–y= 9
8 – (–1) = 9
True
The solution of the system is (8, –1).
Example
Solve the following system of equations
6x – 3y = –3
4x + 5y = –9
Multiply both sides of the first equation by 5 and the second
equation by 3.
5(6x – 3y) = 5(–3)
3(4x + 5y) = 3(–9)
30x – 15y = –15
12x + 15y = –27
42x
= –42
x = –1
Add the equations.
Solve for y.
Divide both sides by 42.
continued
Example (cont)
Substitute –1 for x into one of the original equations
and solve for y.
6x – 3y = –3
6(–1) – 3y = –3
Replace the x value in the first equation.
–6 – 3y = –3 Simplify.
–3y = 3
Add 6 to both sides.
y = –1 Divide both sides by –3.
Our computations have produced the point (–1, –1).
continued
Example (cont)
Check the point in the original equations.
First equation,
6x – 3y = –3
6(–1) – 3(–1) = –3
True
Second equation,
4x + 5y = –9
4(–1) + 5(–1) = –9
True
The solution of the system is (–1, –1).
The Elimination Method
Solving a System of Two Linear Equations Using the
Elimination Method
Step 1: Rewrite each equation in standard form, Ax + By = C.
Step 2: If necessary, multiply one or both equations by some
nonzero number so that the coefficients of a variable
are opposites of each other.
Step 3: Add the equations.
Step 4: Find the value of one variable by solving the equation from
Step 3.
Step 5: Find the value of the second variable by substituting the
value found in Step 4 into either of the original equations.
Step 6: Check the proposed solution in both original equations.
Example
Solve the system of equations using the elimination method.
2
1
3
x y 
3
4
2
1
1
x  y  2
2
4
First multiply both sides of the equations by a number that
will clear the fractions out of the equations.
continued
Example (cont)
Multiply both sides of each equation by 12. (Note: You do
not have to multiply each equation by the same number, but
in this case it will be convenient to do so.)
First equation,
2
1
3
x y 
3
4
2
2 1 
 3
12 x  y   12  
4 
3
 2
8x  3y  18
Multiply both sides by 12.
Simplify.
continued
Example (cont)
Second equation,
1
1
x  y  2
2
4
1 
1
12 x  y   12 2
4 
2
6 x  3 y  24
Multiply both sides by 12.
Simplify.
Add the two resulting equations.
8x + 3y = – 18
6x – 3y = – 24
14x
= – 42
x = –3
Divide both sides by 14.
continued
Example (cont)
Substitute –3 for x into one of the original equations.
8x + 3y = –18
8(–3) + 3y = –18
–24 + 3y = –18
3y = –18 + 24 = 6
y=2
We need to check the ordered pair (–3, 2) in both
equations of the original system.
continued
Example (cont)
Check the point in the original equations.
First equation:
Second equation:
2
1
3
x y 
3
4
2
1
1
x  y  2
2
4
2
1
3
(3)  (2)  
3
4
2
1
1
(3)  (2)  2
2
4
1
3
2  
2
2
3 1
   2
2 2
True
The solution is (–3, 2).
True
Special Cases
In a similar fashion to what you found in the last section,
use of the addition method to combine two equations
might lead you to results like …
5 = 5 (which is always true, thus indicating that there
are infinitely many solutions, since the two equations
represent the same line), or
0 = 6 (which is never true, thus indicating that there
are no solutions, since the two equations represent
parallel lines).