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The Humongous Book of Algebra Problems by W. Michael Kelley

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Chapter Four — Linear Equations in One Variable
The Humongous Book of Algebra Problems
61
4.14 Solve the equation for x.
Before you can isolate x, it must appear in the numerator of the fraction. Take
the reciprocal of both sides of the equation to accomplish thisit will have no
impact on the equality statement.
Write the fraction using an explicit coefficient: .
To eliminate the coefficient , and thereby isolate x, multiply by its reciprocal:
.
Solving Equations Using Multiple Steps
Nothing new here, just more steps
4.15 Solve the equation 4x + 3 = 27 for x.
Subtract 3 from both sides of the equation so that only the variable term 4x is
left of the equal sign and the constants are right of the equal sign.
To eliminate the coefficient of 4x, and thereby solve for x, divide both sides of
the equation by the coefficient.
The
statement
is true,
and it stays true
if you ip both
fractions upside
down:
. After
you learn cross
multiplication in
Problems 21.1–21.8, it
wont matter if the
x is in the numerator
or denominator. For
now, though, youre
solving for x, not
,
so that x needs
to be up top.
Keep anything
with an x in it on
the left side of the
equation (because youre
solving for x). Any term
without an x (such as 3
in this equation) gets
moved to the right
side.
Chapter Four — Linear Equations in One Variable
The Humongous Book of Algebra Problems
62
Note: Problems 4.164.17 refer to the equation 19 – (x – 5) = 4.
4.16 Solve the equation by isolating x on the right side of the equal sign.
Apply the distributive property to simplify the left side of the equation.
You are directed to isolate x right of the equal sign, so add x to both sides of the
equation to eliminate variable terms on the left side.
24 = 4 + x
Subtract 4 from both sides of the equation to solve for x.
Note: Problems 4.164.17 refer to the equation 19 – (x – 5) = 4.
4.17 Solve the equation by isolating x on the left side of the equal sign and verify
that the solutions to Problems 4.16 and 4.17 are equal.
Simplify the left side of the equation.
You are instructed to isolate x on the left side of the equation, so eliminate the
constant left of the equal sign by subtracting it from both sides of the equation.
The equation is not solved for x, because x has a coefficient of –1. Eliminate the
coefficient by multiplying both sides of the equation by –1.
Problems 4.16 and 4.17 have the same solution: x = 20.
4.18 Solve the equation 6x – 3 = 9x + 2 for x.
Group the xterms left of the equal sign by subtracting 9x from both sides of
the equation.
Group the constant terms right of the equal sign by adding 3 to both sides of
the equation.
Pretend
the negative
sign outside (x – 5)
is a –1 and multiply
x and –5 by –1.
You could
also divide both
sides by –1.
Chapter Four — Linear Equations in One Variable
The Humongous Book of Algebra Problems
63
Divide both sides of the equation by the coefficient of x.
4.19 Solve the equation for x.
Move the x-terms to the numerators of the fractions by taking the reciprocal of
both sides of the equation.
Write each fraction as the product of a rational number and x.
Move the variable terms left of the equal sign by subtracting from both
sides of the equation.
Use the least common denominator 15 to combine the like terms.
Multiply both sides of the equation by the reciprocal of .
Although x = 0 appears to be a valid solution, it is not. Substituting x = 0 into the
original equation produces undefined statements, therefore invalidating the
solution.
Dont for-
get that
you can always
check your answer
by plugging it back
into the original
equation for x:
Both of
these terms have
the same variable
(x), so theyre like
terms. Combine their
coefcients and
multiply the result
by the common
variable x.

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