Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
162
Variable Elimination
Make one variable disappear and solve for the other one
8.19 Explain how to solve a system of two linear equations in two variables using the
variable elimination technique.
To apply the variable elimination method, the coefficients of either the x-terms
or the y-terms of the equations in the system must be opposites. Therefore,
when the equations are added, a new equation is generated that contains only
one variable. In many cases, you must multiply one or both of the equations by a
nonzero real number to introduce the coefficient opposites into the system.
Note: Problems 8.20–8.21 refer to the following system of equations.
8.20 Solve the system by eliminating y.
The equations of the system have opposite y-coefficients—in the first equation,
y has a coefficient of 1, and in the second, the y-coefficient is –1. Add the
equations of the system together by combining like terms.
The result is a linear equation in one variable: 3x = 9. Solve it for x.
Substitute x into either equation of the system to determine the corresponding
value of y.
The solution to the system is (x,y) = (3,1).
The book
substituted into
the rst equation of
the system, but you
get the same thing
when you substitute
x = 3 into the second
equation:
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
163
Note: Problems 8.20–8.21 refer to the following system of equations.
8.21 Solve the system by eliminating x.
To eliminate the x-variable from the system, the x-coefficients of the equations
must be opposites. Multiplying the terms of the second equation by –2 achieves
this goal.
Add the equations of the modified system to eliminate y.
Solve the equation 3y = 3 for y.
Substitute y = 1 into either equation of the system to determine the
corresponding value of x.
The solution to the system is (x,y) = (3,1).
You could
multiply the
rst equation by
instead, making
its x-coefcient –1,
the opposite of the x-
coefcient in the other
equation. However,
that’s not a great
idea, because it turns
the y-coefcient
and the constant in
the rst equation
into fractions,
and nobody likes
fractions.
This is the same
solution as Problem
8.20, so it doesn’t
matter which variable
you eliminate—you end
up with the same nal
answer.
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
164
Note: Problems 8.22–8.23 refer to the system of equations below.
8.22 Solve the system by eliminating x.
The coefficients of the x-terms in the system are 1 and 5. The easiest way to
transform those numbers into opposites is to multiply the terms of the first
equation by –5.
Add the equations of the modified system and solve the resulting equation for y.
Substitute y = –9 into either equation of the original system to calculate the
corresponding value of x.
The solution to the system is (x,y) = (–2,–9).
Note: Problems 8.22–8.23 refer to the system of equations below.
8.23 Solve the system by eliminating y.
Multiplying the first equation by 3 changes its y-coefficient to the opposite of
the other y-coefficient in the system. Add the equations together and solve for x.
There’s
lots of
things you
COULD do. For
instance, you
could multiply the
rst equation by
10 and the second
by –2 and the x-
coefcients would be
opposites: 1
˙
10 = 10
and 5
˙
(–2) = –10.
Just make sure you
multiply ALL the
terms in both
equations by
the numbers
you choose.
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
165
Substitute x = –2 into either equation of the system to complete the solution.
The solution to the system is (x,y) = (–2,–9).
8.24 Problem 8.14 used the substitution method to solve the following system of
equations. Verify its solution using variable elimination.
Multiplying the second equation by –2 makes the x-coefficients of the system
into opposites. Add the equations of the modified system and solve for y.
Substitute into either equation of the system to calculate the
corresponding value of x.
The solution to the system is , the same solution calculated in
Problem 8.14.
You could
multiply the
second equation
by 3 and turn the y-
coefcients opposites:
–12 and 12. If you’re
feeling crazy, you could
even multiply the rst
equation by 5 and the
second by –10 to turn
the x-coefcients
into 2
˙
5 = 10 and
1
˙
(–10) = –10.
This way
was much
faster. Substitution
usually requires more
steps than variable
elimination.
Chapter Eight — Systems of Linear Equations and Inequalities
The Humongous Book of Algebra Problems
166
8.25 Problem 8.15 used the substitution method to solve the below system of
equations. Verify its solution using variable elimination.
To eliminate the y-terms, multiply the first equation by 2 and the second by 3.
Add the equations and solve for x.
Substitute x = 6 into either equation of the system to calculate the
corresponding y-value.
The solution to the system is (x,y) = (6,11), the same solution calculated in
Problem 6.15.
8.26 Problem 8.16 used the substitution method to solve the below system of
equations. Verify its solution using variable elimination.
Multiply the second equation by 3 to get 27x + 15y = 24, add the equations of the
modified system, and solve for x.
You could
do the same
with the x-terms:
Multiply the rst
equation by 3 and
the second equation
by –7. You’re basically
multiplying one
equation by the x-
coefcient of the
other equation.
You could
multiply the rst
equation by 9 and
the second by –2
(or the rst by –9
and the second by
2) to eliminate the
x’s instead.
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