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Chapter 2

Toeing the Straight Line:
Linear Equations
In This Chapter
Isolating values of x in linear equations
Comparing variable values with inequalities
Assessing absolute value in equations and inequalities


he term linear has the word line buried in it, and the obvious connection
is that you can graph many linear equations as lines. But linear expressions can come in many types of packages, not just equations or lines. Add
an interesting operation or two, put several first-degree terms together, throw
in a funny connective, and you can construct all sorts of creative mathematical challenges. In this chapter, you find out how to deal with linear equations,
what to do with the answers in linear inequalities, and how to rewrite linear
absolute-value equations and inequalities so that you can solve them.

Linear Equations: Handling
the First Degree
Linear equations feature variables that reach only the first degree, meaning
that the highest power of any variable you solve for is one. The general form
of a linear equation with one variable is
ax + b = c
The one variable is the x. (If you go to Chapter 12, you can see linear equations
with two or three variables.) But, no matter how many variables you see, the


Part I: Homing in on Basic Solutions
common theme to linear equations is that each variable has only one solution or value that works in the equation.
The graph of the single solution, if you really want to graph it, is one point on
the number line — the answer to the equation. When you up the ante to two
variables in a linear equation, the graph of all the solutions (there are infinitely
many) is a straight line. Any point on the line is a solution. Three variables
means you have a plane — a flat surface.
Generally, algebra uses the letters at the end of the alphabet for variables;
the letters at the beginning of the alphabet are reserved for coefficients and

Tackling basic linear equations
To solve a linear equation, you isolate the variable on one side of the equation by adding the same number to both sides — or you can subtract, multiply, or divide the same number on both sides.
For example, you solve the equation 4x – 7 = 21 by adding 7 to each side of
the equation, to isolate the variable and the multiplier, and then dividing
each side by 4, to leave the variable on its own:
4x –7 + 7 = 21 + 7 → 4x = 28
4x ÷ 4 = 28 ÷ 4 → x = 7
When a linear equation has grouping symbols such as parentheses, brackets,
or braces, you deal with any distributing across and simplifying within the
grouping symbols before you isolate the variable. For instance, to solve the
equation 5x – [3(x + 2) – 4(5 – 2x) + 6] = 20, you first distribute the 3 and –4
inside the brackets:
5x - 8 3 ^ x + 2h - 4 ^ 5 - 2x h + 6 B = 20
5x - 6 3x + 6 - 20 + 8x + 6 @ = 20
You then combine the terms that combine and distribute the negative sign
(–) in front of the bracket; it’s like multiplying through by –1:
5x - 611x - 8 @ = 20
5x - 11x + 8 = 20
Simplify again, and you can solve for x:

Chapter 2: Toeing the Straight Line: Linear Equations
- 6x + 8 = 20
- 6x = 12
x = -2
When distributing a number or negative sign over terms within a grouping
symbol, make sure you multiply every term by that value or sign. If you don’t
multiply each and every term, the new expression won’t be equivalent to the
To check your answer from the previous example problem, replace every x
in the original equation with –2. If you do so, you get a true statement. In this
case, you get 20 = 20. The solution –2 is the only answer that works — focusing
your work on just one answer is what’s nice about linear equations.

Clearing out fractions
The problem with fractions, like cats, is that they aren’t particularly easy
to deal with. They always insist on having their own way — in the form of
common denominators before you can add or subtract. And division? Don’t
get me started!
Seriously, though, the best way to deal with linear equations that involve
variables tangled up with fractions is to get rid of the fractions. Your game
plan is to multiply both sides of the equation by the least common denominator of all the fractions in the equation.
To solve x + 2 + 4x + 2 = 9 - x , for example, you multiply each term in the
equation by 70 — the least common denominator (also known as the least
common multiple) for fractions with the denominators 5, 7, and 2:
N 10 J
N 35 J
70 KK x + 2 OO + 70 KK 4x + 2 OO = 70 KK 9 - x OO
L 1 P
L 1 P


Now you distribute the reduced numbers over each parenthesis, combine the
like terms, and solve for x:
14 ^ x + 2h + 10 ^ 4x + 2h = 35 ^ 9 - x h
14x + 28 + 40x + 20 = 315 - 35x
54x + 48 = 315 - 35x
89x = 267



Part I: Homing in on Basic Solutions
Extraneous (false) solutions can occur when you alter the original format of an
equation. When working with fractions and changing the form of an equation to
a more easily solved form, always check your answer in the original equation.
For the previous example problem, you insert x = 3 into x + 2 + 4x + 2 = 9 - x
and get 3 = 3.

Isolating different unknowns
When you see only one variable in an equation, you have a pretty clear
idea what you’re solving for. When you have an equation like 4x + 2 = 11 or
5(3z – 11) + 4z = 15(8 + z), you identify the one variable and start solving for it.
Life isn’t always as easy as one-variable equations, however. Being able to
solve an equation for some variable when it contains more than one unknown
can be helpful in many situations. If you’re repeating a task over and over —
such as trying different widths of gardens or diameters of pools to find the
best size — you can solve for one of the variables in the equation in terms of
the others.
The equation A = 1⁄2h(b1 + b2), for example, is the formula you use to find the
area of a trapezoid. The letter A represents area, h stands for height (the distance between the two parallel bases), and the two b’s are the two parallel
sides called the bases of the trapezoid.
If you want to construct a trapezoid that has a set area, you need to figure
out what dimensions give you that area. You’ll find it easier to do the many
computations if you solve for one of the components of the formula first —
for h, b1, or b2.
To solve for h in terms of the rest of the unknowns or letters, you multiply
each side by two, which clears out the fraction, and then divide by the entire
expression in the parenthesis:
A = 1 h _b 1 + b 2i
2A = 2 $ 1 h _ b 1 + b 2 i
h _b 1 + b 2i
_b 1 + b 2i
_b 1 + b 2i

_b 1 + b 2i


Chapter 2: Toeing the Straight Line: Linear Equations

Paying off your mortgage with algebra
A few years ago, one of my mathematically challenged friends asked me if I could help her figure
out what would happen to her house payments
if she paid $100 more each month on her mortgage. She knew that she’d pay off her house
faster, and she’d pay less in interest. But how
long would it take and how much would she
save? I created a spreadsheet and used the formula for an amortized loan (mortgage). I made
different columns showing the principal balance

that remained (solved for P) and the amount of
the payment going toward interest (solved for the
difference), and I extended the spreadsheet
down for the number of months of the loan. We
put the different payment amounts into the original formula to see how they changed the total
number of payments and the total amount paid.
She was amazed. I was even amazed! She’s
paying off her mortgage much sooner than

You can also solve for b2, the measure of the longer base of the trapezoid. To
do so, you multiply each side of the equation by two, divide each side by h,
and then subtract b1 from each side:
A = 1 h _b 1 + b 2i
2A = 2 $ 1 h _ b 1 + b 2 i

2A = h _ b 1 + b 2 i
2A = b + b
2A - b = b

You can leave the equation in that form, with two terms, or you can find a
common denominator and combine the terms on the left:
2A - b 1 h
= b2
When you rewrite a formula aimed at solving for a particular unknown, you can
put the formula into a graphing calculator or spreadsheet to do some investigating into how changes in the individual values change the variable that you
solve for (see a spreadsheet example of this in the “Paying off your mortgage
with algebra” sidebar).



Part I: Homing in on Basic Solutions

Linear Inequalities: Algebraic
Relationship Therapy
Equations — statements with equal signs — are one type of relationship or
comparison between things; they say that terms, expressions, or other entities
are exactly the same. An inequality is a bit less precise. Algebraic inequalities
show relationships between a number and an expression or between two
expressions. In other words, you use inequalities for comparisons.
Inequalities in algebra are less than (<), greater than (>), less than or equal to
(≤), and greater than or equal to (≥). A linear equation has only one solution,
but a linear inequality has an infinite number of solutions. When you write
x ≤ 7, for example, you can replace x with 6, 5, 4, –3, –100, and so on, including all the fractions that fall between the integers that work in the inequality.
Here are the rules for operating on inequalities (you can replace the < symbol
with any of the inequality symbols, and the rule will still hold):
If a < b, a + c < b + c (adding any number).
If a < b, a – c < b – c (subtracting any number).
If a < b, a ⋅ c < b ⋅ c (multiplying by any positive number).
If a < b, a ⋅ c > b ⋅ c (multiplying by any negative number).
If a < b, a
c < c (dividing by any positive number).
If a < b, a
c > c (dividing by any negative number).
b c d
If a
c < d , a > b (reciprocating fractions).
You must not multiply or divide each side of an inequality by zero. If you do
so, you create an incorrect statement. Multiplying each side of 3 < 4 by 0, you
get 0 < 0, which is clearly a false statement. You can’t divide each side by 0,
because you can never divide anything by 0 — no such number with 0 in the
denominator exists.

Solving basic inequalities
To solve a basic inequality, you first move all the variable terms to one side
of the inequality and the numbers to the other. After you simplify the inequality down to a variable and a number, you can find out what values of the
variable will make the inequality into a true statement. For example, to solve
3x + 4 > 11 – 4x, you add 4x to each side and subtract 4 from each side. The

Chapter 2: Toeing the Straight Line: Linear Equations
inequality sign stays the same because no multiplication or division by negative numbers is involved. Now you have 7x > 7. Dividing each side by 7 also
leaves the sense (direction of the inequality) untouched because 7 is a positive number. Your final solution is x > 1. The answer says that any number
larger than one can replace the x’s in the original inequality and make the
inequality into a true statement.
The rules for solving linear equations (see the section “Linear Equations:
Handling the First Degree”) also work with inequalities — somewhat. Everything goes smoothly until you try to multiply or divide each side of an inequality by a negative number.
When you multiply or divide each side of an inequality by a negative number,
you have to reverse the sense (change < to >, or vice versa) to keep the inequality true.
The inequality 4(x – 3) – 2 ≥ 3(2x + 1) + 7, for example, has grouping symbols
that you have to deal with. Distribute the 4 and 3 through their respective
multipliers to make the inequality into 4x – 12 – 2 ≥ 6x + 3 + 7. Simplify the
terms on each side to get 4x – 14 ≥ 6x + 10. Now you put your inequality skills
to work. Subtract 6x from each side and add 14 to each side; the inequality
becomes –2x ≥ 24. When you divide each side by –2, you have to reverse the
sense; you get the answer x ≤ –12. Only numbers smaller than –12 or exactly
equal to –12 work in the original inequality.
When solving the previous example, you have two choices when you get to
the step 4x – 14 ≥ 6x + 10, based on the fact that the inequality a < b is equivalent to b > a. If you subtract 6x from both sides, you end up dividing by a negative number. If you move the variables to the right and the numbers to the
left, you don’t have to divide by a negative number, but the answer looks a
bit different. If you subtract 4x from each side and subtract 10 from each
side, you get –24 ≥ 2x. When you divide each side by 2, you don’t change the
sense, and you get –12 ≥ x. You read the answer as “–12 is greater than or
equal to x.” This inequality has the same solutions as x ≤ –12, but stating the
inequality with the number coming first is a bit more awkward.

Introducing interval notation
You can alleviate the awkwardness of writing answers with inequality notation by using another format called interval notation. You use interval notation extensively in calculus, where you’re constantly looking at different
intervals involving the same function. Much of higher mathematics uses
interval notation, although I really suspect that book publishers pushed its
use because it’s quicker and neater than inequality notation. Interval notation uses parentheses, brackets, commas, and the infinity symbol to bring
clarity to the murky inequality waters.



Part I: Homing in on Basic Solutions
And, surprise surprise, the interval-notation system has some rules:
You order any numbers used in the notation with the smaller number to
the left of the larger number.
You indicate “or equal to” by using a bracket.
If the solution doesn’t include the end number, you use a parenthesis.
When the interval doesn’t end (it goes up to positive infinity or down
to negative infinity), use +∞ or –∞, whichever is appropriate, and a
Here are some examples of inequality notation and the corresponding interval notation:
x < 3 → (–∞, 3)
x ≥ –2 → [–2, ∞)
4 ≤ x < 9 → [4, 9)
–3 < x < 7 → (–3, 7)
Notice that the second example has a bracket by the –2, because the “greater
than or equal to” indicates that you include the –2, also. The same is true of
the 4 in the third example. The last example shows you why interval notation
can be a problem at times. Taken out of context, how do you know if (–3, 7)
represents the interval containing all the numbers between –3 and 7 or if it
represents the point (–3, 7) on the coordinate plane? You can’t tell. A problem
containing such notation has to give you some sort of hint.

Compounding inequality issues
A compound inequality is an inequality with more than one comparison or
inequality symbol — for instance, –2 < x ≤ 5. To solve compound inequalities
for the value of the variables, you use the same inequality rules (see the intro
to this section), and you expand the rules to apply to each section (intervals
separated by inequality symbols).
To solve the inequality –8 ≤ 3x – 5 < 10, for example, you add 5 to each of the
three sections and then divide each section by 3:
- 8 # 3x - 5 < 10
+5 +5
- 3 # 3x
< 15
- 3 # 3x
< 15
-1 # x

Chapter 2: Toeing the Straight Line: Linear Equations

Ancient symbols for timeless operations
Many ancient cultures used their own symbols
for mathematical operations, and the cultures
that followed altered or modernized the symbols
for their own use. You can see one of the first
symbols used for addition in the following figure,
located on the far left — a version of the Italian
capital P for the word piu, meaning plus.
Tartaglia, a self-taught 16th century Italian
mathematician, used this symbol for addition
regularly. The modern plus symbol, +, is probably a shortened form of the Latin word et, meaning and.

The second figure from the left is what Greek
mathematician Diophantes liked to use in ancient

Greek times for subtraction. The modern subtraction symbol, –, may be a leftover from what
the traders in medieval times used to indicate differences in product weights.
Leibniz, a child prodigy from the 17th century
who taught himself Latin, preferred the third
symbol from the left for multiplication. One
modern multiplication symbol, x or ⋅, is based on
St. Andrew’s Cross, but Leibniz used the open
circle because he thought that the modern
symbol looked too much like the unknown x.
The symbol on the far right is a somewhat backward D, used in the 18th century by French mathematician Gallimard for division. The modern
division symbol, ÷, may come from a fraction line
with dots added above and below.

You write the answer, –1 ≤ x < 5, in interval notation as [–1, 5).
Here’s a more complicated example. You solve the problem –1 < 5 – 2x ≤ 7 by
subtracting 5 from each section and then dividing each section by –2. Of
course, dividing by a negative means that you turn the senses around:
- 1 < 5 - 2x # 7
- 5 -5
-6 <
- 2x # 2
- 6 > - 2x $ 2
3> x
$ -1
You write the answer, 3 > x ≥ –1, backward as far as the order of the numbers
on the number line; the number –1 is smaller than 3. To flip the inequality in
the opposite direction, you reverse the inequalities, too: –1 ≤ x < 3. In interval
notation, you write the answer as [–1, 3).


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