Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
158
8.15 Solve the following system using substitution.
None of the variables in this system has a coefficient of 1, so none distinguishes
itself as the preferred variable to isolate and substitute into the other equation
of the system. In these cases, it is usually preferable to solve for the variable
whose coefficient has the least absolute value. Therefore, you should solve
3x + 2y = 40 for y.
Substitute into the first equation of the system.
Multiply each of the terms by 2 to eliminate the fraction, and solve for x.
Complete the solution by substituting x = 6 into the equation you solved for y.
The solution to the system is (x,y) = (6,11).
Ignore the
signs of the
coefcients and pick
the smallest number.
Here, the coefcients
are 2, 3, and 7. The
smallest of those is 2,
from the term 2y in
the second equation,
so solve the second
equation for y.
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
159
8.16 Solve the below system using substitution.
Solve 2x – 15y = 5 for x, as it has the coefficient with the smallest absolute value.
Substitute into the second equation.
Multiply each of the terms by 2 to eliminate the fractions, and solve for y.
Substitute into , the equation you solved for x.
The solution to the system is .
Problem
8.15 gave
this advice:
When none of
the coefcients
is 1, then solve
for the coefcient
that’s smallest when
you ignore the signs.
Sure –15 < 2, but
when you ignore the
signs, 2 < 15, so 2
is the smallest
coefcient.
If you
plug it into
the FIRST
equation,
everything cancels
out and you get
0 = 0. That’s
because you’re
actually plugging
another version of
the rst equation
into itself. You have
to substitute the
equation you
solved for x into
the OTHER
equation of
the system.
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
160
8.17 Solve the following system using substitution.
The first equation of this system is written in terms of three variables, and the
second is written in terms of two variables. Because the third equation is written
in terms of a single variable, it should be solved first.
Of the two remaining equations in the system, 8y – 4z = 5 has fewer variables.
In fact, substituting into the equation leaves a single variable, y, for
which you can solve.
Now substitute and into the first equation of the system and solve
it for x to complete the solution.
The solution to the system is .
8.18 Solve the below system using substitution.
Unlike Problem 8.17, none of the equations in this system is written in terms
of a single variable. However, all the equations contain one common variable:
y. Solve the third equation for y to get y = 6z + 6 and substitute it into the other
equations of the system.
Substitution
isn’t restricted
to systems with
two equations. When
there are three or
more, start with the
equations that have
the fewest variables
in them, such as
4z + 1 = 0 in this
system.
Three
different
variables are
in the system,
even though they
all don’t appear
in each of the
equations. That
means the solution
to the system
must have all
three: (x,y,z).
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
161
The end result of substituting y into the first two equations of the system is a
new system of two linear equations in two variables.
Solve this system of equations using substitution: solve the second equation for x
to get x = 27z + 29 and substitute that into the first equation.
Substitute z = –1 into the equation you solved for x.
Finally, substitute z = –1 into the equation you originally solved for y to complete
the solution.
The solution to the system is (x,y,z) = (2,0,–1).
Dividing every-
thing by 2
doesn’t change
the solution to the
equation, and the
added benet is
that x now has a
coefcient of 1, so
you can solve for
it easily.
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