Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
153
Figure 8-5: The graphs of and x – 3y = –9 both have slope and are,
therefore, parallel lines.
By definition, parallel lines do not intersect in the coordinate plane. It is
the intersection point of a system of equations that defines its solution, so the
absence of an intersection point indicates the absence of a solution to the
system. Systems of equations with no solutions are classified as “inconsistent.”
The Substitution Method
Solve one equation for a variable and plug it into the other
8.8 Explain how to apply the substitution method to solve a system of two linear
equations in two variables.
If one equation of a system can easily be solved for one of its variables, you can
substitute the resulting expression into the other equation of the system. For
instance, if the first equation of a system can be solved for y, do so, and then
substitute the result for y in the second equation.
The net result is a new equation written in terms of a single variable, in this case
x. When solved, it produces the x-value of the solution to the system, which can
then be substituted into one of the equations of the system to determine the
corresponding y-value.
To prove
that there
is no solution
to a system of
linear equations,
all you have to
do is prove that
all the lines have
the same slope (or
that all the lines
have NO slope—
vertical lines
are parallel
as well).
You’ll learn
a couple of ways
to solve systems of
equations. This one
works best if you can
easily solve one of the
equations for x or y.
It’s even easier when
an equation is ALREADY
solved for x or y, like in
Problems 8.9 and 8.10.
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
154
Note: Problems 8.9–8.10 refer to the following system of equations.
8.9 Solve the system using substitution.
The second equation in the system is solved for y: y = 11x – 16. According to this
statement, the expressions y and 11x – 16 have the same value in this system of
equations. Therefore, you may substitute 11x – 16 for y in the other equation of
the system, x + y = 8.
Solve the equation for x.
The solution to a system of linear equations in two variables is a coordinate
pair (x,y). To determine the y-coordinate, and therefore complete the solution,
substitute x = 2 into the equation y = 11x – 16.
The solution to the system of equations is (x,y) = (2,6).
Note: Problems 8.9–8.10 refer to the following system of equations.
8.10 Verify the solution generated in Problem 8.9.
According to Problem 8.9, the solution to the system of equations is
(x,y) = (2,6). To verify that the solution is correct, substitute x = 2 and y = 6
into both equations of the system.
Substituting (x,y) = (2,6) into the equations of the system produces true
statements, so it is the correct solution to the system.
You could
substitute it
into the other
equation, x + y = 8,
instead. You’ll get
the same answer.
However, it’s almost
always quickest
to plug x into the
equation already
solved for y (and
vice versa).
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
155
8.11 Solve the following system of equations using substitution.
According to the first equation, the expressions x and have the same value
in this system. Substitute for x in the second equation and solve for y.
Substitute y = 5 into the equation to complete the solution.
The solution to the system is (x,y) = (–3,5).
Rewrite the
coefcient of y (which
is 1 even though it’s not
written explicitly) using a common
denominator:
The answer
(x,y) = (5,–3) is
NOT correct. When
you write (x,y), you’re
saying that the rst
value is x and the
second value is y, and
in this problem x = –3
and y = 5, not
the other way
around.
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
156
Note: Problems 8.12–8.13 refer to the following system of equations.
8.12 Solve the system using substitution.
Neither of the equations is solved for x or y, but solving 6x + y = 1 for y is a simple
matter. Subtract 6x from both sides of the equation to get y = –6x + 1. Substitute
this expression for y in the other equation of the system and solve for x.
Substitute into the equation you solved for y to complete the solution.
The solution to the system is .
Note: Problems 8.12–8.13 refer to the following system of equations.
8.13 Verify the solution generated in Problem 8.12.
According to Problem 8.12, the solution is . Substitute
and y = –2 into both equations of the system and verify that the results are true
statements.
When a
variable has a
coefcient of 1, like
y in the rst equation
of this system, solving
for that variable is
usually quite easy.
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
157
8.14 Solve the following system using substitution.
Solve the second equation for x to get x = –4y – 4. Substitute this expression into
the first equation and solve for y.
Substitute into either equation of the system and solve for x to complete
the solution.
The solution to the system is .
Just like
in Problem
8.12, the
variable with a
coefcient of 1
is the easiest to
solve for. You can
technically solve for
ANY of the variables
here, and you’ll end
up with the same
solution, but why
work harder
than you have
to?
The book plugs y into the rst equation,
but you get the same x-value when you
plug y into the second equation:
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