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 this—it 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

won’t matter if the

x is in the numerator

or denominator. For

now, though, you’re

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 you’re

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 4.16–4.17 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 4.16–4.17 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 x–terms 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.

Don’t 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 they’re like

terms. Combine their

coefcients and

multiply the result

by the common

variable x.

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