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Source: Elementary algebra

Elementary algebra, also known as college algebra, encompasses the basic concepts of algebra. It is often contrasted with arithmetic: arithmetic deals with specified numbers, whilst algebra introduces variables (quantities without fixed values).
This use of variables entails use of algebraic notation and an understanding of the general rules of the operations introduced in arithmetic: addition, subtraction, multiplication, division, etc. Unlike abstract algebra, elementary algebra is not concerned with algebraic structures outside the realm of real and complex numbers.
It is typically taught to secondary school students and at introductory college level in the United States, and builds on their understanding of arithmetic. The use of variables to denote quantities allows general relationships between quantities to be formally and concisely expressed, and thus enables solving a broader scope of problems. Many quantitative relationships in science and mathematics are expressed as algebraic equations.

Algebraic operations

Algebraic notation

= Alternative notation

Concepts

= Variables

Elementary

algebra

= Simplifying expressions

= Equations

where

<

{\displaystyle <}

represents 'less than'. Just like standard equality equations, numbers can be added, subtracted, multiplied or divided. The only exception is that when multiplying or dividing by a negative number, the inequality symbol must be flipped.

Properties of equality

By definition, equality is an equivalence relation, meaning it is reflexive (i.e.

b
=
b

{\displaystyle b=b}

), symmetric (i.e. if

a
=
b

{\displaystyle a=b}

then

b
=
a

{\displaystyle b=a}

), and transitive (i.e. if

a
=
b

{\displaystyle a=b}

and

b
=
c

{\displaystyle b=c}

then

a
=
c

{\displaystyle a=c}

). It also satisfies the important property that if two symbols are used for equal things, then one symbol can be substituted for the other in any true statement about the first and the statement will remain true. This implies the following properties:

if

a
=
b

{\displaystyle a=b}

and

c
=
d

{\displaystyle c=d}

then

a
+
c
=
b
+
d

{\displaystyle a+c=b+d}

and

a
c
=
b
d

{\displaystyle ac=bd}

;
if

a
=
b

{\displaystyle a=b}

then

a
+
c
=
b
+
c

{\displaystyle a+c=b+c}

and

a
c
=
b
c

{\displaystyle ac=bc}

;
more generally, for any function f, if

a
=
b

{\displaystyle a=b}

then

f
(
a
)
=
f
(
b
)

{\displaystyle f(a)=f(b)}

.

Properties of inequality

The relations less than

<

{\displaystyle <}

and greater than

>

{\displaystyle >}

have the property of transitivity:

If

a
<
b

{\displaystyle a
and

b
<
c

{\displaystyle b
then

a
<
c

{\displaystyle a
;
If

a
<
b

{\displaystyle a
and

c
<
d

{\displaystyle c
then

a
+
c
<
b
+
d

{\displaystyle a+c
;
If

a
<
b

{\displaystyle a
and

c
>
0

{\displaystyle c>0}

then

a
c
<
b
c

{\displaystyle ac
;
If

a
<
b

{\displaystyle a
and

c
<
0

{\displaystyle c<0}

then

b
c
<
a
c

{\displaystyle bc
.
By reversing the inequation,

<

{\displaystyle <}

and

>

{\displaystyle >}

can be swapped, for example:

a
<
b

{\displaystyle a
is equivalent to

b
>
a

{\displaystyle b>a}

= Substitution

=

Substitution is replacing the terms in an expression to create a new expression. Substituting 3 for a in the expression a*5 makes a new expression 3*5 with meaning 15. Substituting the terms of a statement makes a new statement. When the original statement is true independently of the values of the terms, the statement created by substitutions is also true. Hence, definitions can be made in symbolic terms and interpreted through substitution: if

a

2

:=
a
×
a

{\displaystyle a^{2}:=a\times a}

is meant as the definition of

a

2

,

{\displaystyle a^{2},}

as the product of a with itself, substituting 3 for a informs the reader of this statement that

3

2

{\displaystyle 3^{2}}

means 3 × 3 = 9. Often it's not known whether the statement is true independently of the values of the terms. And, substitution allows one to derive restrictions on the possible values, or show what conditions the statement holds under. For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1.
If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0. Then we can substitute again, letting x = b and y = c, to show that if bc = 0 then b = 0 or c = 0. Therefore, if abc = 0, then a = 0 or (b = 0 or c = 0), so abc = 0 implies a = 0 or b = 0 or c = 0.
If the original fact were stated as "ab = 0 implies a = 0 or b = 0", then when saying "consider abc = 0," we would have a conflict of terms when substituting. Yet the above logic is still valid to show that if abc = 0 then a = 0 or b = 0 or c = 0 if, instead of letting a = a and b = bc, one substitutes a for a and b for bc (and with bc = 0, substituting b for a and c for b). This shows that substituting for the terms in a statement isn't always the same as letting the terms from the statement equal the substituted terms. In this situation it's clear that if we substitute an expression a into the a term of the original equation, the a substituted does not refer to the a in the statement "ab = 0 implies a = 0 or b = 0."

Solving algebraic equations

The following sections lay out examples of some of the types of algebraic equations that may be encountered.

= Linear equations with one variable

=

Linear equations are so-called, because when they are plotted, they describe a straight line. The simplest equations to solve are linear equations that have only one variable. They contain only constant numbers and a single variable without an exponent. As an example, consider:

Problem in words: If you double the age of a child and add 4, the resulting answer is 12. How old is the child?
Equivalent equation:

2
x
+
4
=
12

{\displaystyle 2x+4=12}

where x represent the child's age
To solve this kind of equation, the technique is add, subtract, multiply, or divide both sides of the equation by the same number in order to isolate the variable on one side of the equation. Once the variable is isolated, the other side of the equation is the value of the variable. This problem and its solution are as follows:

In words: the child is 4 years old.
The general form of a linear equation with one variable, can be written as:

a
x
+
b
=
c

{\displaystyle ax+b=c}

Following the same procedure (i.e. subtract b from both sides, and then divide by a), the general solution is given by

x
=

c
−
b

a

{\displaystyle x={\frac {c-b}{a}}}

= Linear equations with two variables

=

A linear equation with two variables has many (i.e. an infinite number of) solutions. For example:

Problem in words: A father is 22 years older than his son. How old are they?
Equivalent equation:

y
=
x
+
22

{\displaystyle y=x+22}

where y is the father's age, x is the son's age.
That cannot be worked out by itself. If the son's age was made known, then there would no longer be two unknowns (variables). The problem then becomes a linear equation with just one variable, that can be solved as described above.
To solve a linear equation with two variables (unknowns), requires two related equations. For example, if it was also revealed that:

Problem in words
In 10 years, the father will be twice as old as his son.
Equivalent equation

y
+
10

=
2
×
(
x
+
10
)

y

=
2
×
(
x
+
10
)
−
10

Subtract 10 from both sides

y

=
2
x
+
20
−
10

Multiple out brackets

y

=
2
x
+
10

Simplify

{\displaystyle {\begin{aligned}y+10&=2\times (x+10)\\y&=2\times (x+10)-10&&{\text{Subtract 10 from both sides}}\\y&=2x+20-10&&{\text{Multiple out brackets}}\\y&=2x+10&&{\text{Simplify}}\end{aligned}}}

Now there are two related linear equations, each with two unknowns, which enables the production of a linear equation with just one variable, by subtracting one from the other (called the elimination method):

{

y
=
x
+
22

First equation

y
=
2
x
+
10

Second equation

{\displaystyle {\begin{cases}y=x+22&{\text{First equation}}\\y=2x+10&{\text{Second equation}}\end{cases}}}

Subtract the first equation from

(
y
−
y
)

=
(
2
x
−
x
)
+
10
−
22

the second in order to remove

y

0

=
x
−
12

Simplify

12

=
x

Add 12 to both sides

x

=
12

Rearrange

{\displaystyle {\begin{aligned}&&&{\text{Subtract the first equation from}}\\(y-y)&=(2x-x)+10-22&&{\text{the second in order to remove }}y\\0&=x-12&&{\text{Simplify}}\\12&=x&&{\text{Add 12 to both sides}}\\x&=12&&{\text{Rearrange}}\end{aligned}}}

In other words, the son is aged 12, and since the father 22 years older, he must be 34. In 10 years, the son will be 22, and the father will be twice his age, 44. This problem is illustrated on the associated plot of the equations.
For other ways to solve this kind of equations, see below, System of linear equations.

= Quadratic equations

=

A quadratic equation is one which includes a term with an exponent of 2, for example,

x

2

{\displaystyle x^{2}}

, and no term with higher exponent. The name derives from the Latin quadrus, meaning square. In general, a quadratic equation can be expressed in the form

a

x

2

+
b
x
+
c
=
0

{\displaystyle ax^{2}+bx+c=0}

, where a is not zero (if it were zero, then the equation would not be quadratic but linear). Because of this a quadratic equation must contain the term

a

x

2

{\displaystyle ax^{2}}

, which is known as the quadratic term. Hence

a
≠
0

{\displaystyle a\neq 0}

, and so we may divide by a and rearrange the equation into the standard form

x

2

+
p
x
+
q
=
0

{\displaystyle x^{2}+px+q=0}

where

p
=

b
a

{\displaystyle p={\frac {b}{a}}}

and

q
=

c
a

{\displaystyle q={\frac {c}{a}}}

. Solving this, by a process known as completing the square, leads to the quadratic formula

x
=

−
b
±

b

2

−
4
a
c

2
a

,

{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}},}

where the symbol "±" indicates that both

x
=

−
b
+

b

2

−
4
a
c

2
a

and

x
=

−
b
−

b

2

−
4
a
c

2
a

{\displaystyle x={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\quad {\text{and}}\quad x={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}}

are solutions of the quadratic equation.
Quadratic equations can also be solved using factorization (the reverse process of which is expansion, but for two linear terms is sometimes denoted foiling). As an example of factoring:

x

2

+
3
x
−
10
=
0
,

{\displaystyle x^{2}+3x-10=0,}

which is the same thing as

(
x
+
5
)
(
x
−
2
)
=
0.

{\displaystyle (x+5)(x-2)=0.}

It follows from the zero-product property that either

x
=
2

{\displaystyle x=2}

or

x
=
−
5

{\displaystyle x=-5}

are the solutions, since precisely one of the factors must be equal to zero. All quadratic equations will have two solutions in the complex number system, but need not have any in the real number system. For example,

x

2

+
1
=
0

{\displaystyle x^{2}+1=0}

has no real number solution since no real number squared equals −1.
Sometimes a quadratic equation has a root of multiplicity 2, such as:

(
x
+
1

)

2

=
0.

{\displaystyle (x+1)^{2}=0.}

For this equation, −1 is a root of multiplicity 2. This means −1 appears twice, since the equation can be rewritten in factored form as

[
x
−
(
−
1
)
]
[
x
−
(
−
1
)
]
=
0.

{\displaystyle [x-(-1)][x-(-1)]=0.}

Complex numbers

All quadratic equations have exactly two solutions in complex numbers (but they may be equal to each other), a category that includes real numbers, imaginary numbers, and sums of real and imaginary numbers. Complex numbers first arise in the teaching of quadratic equations and the quadratic formula. For example, the quadratic equation

x

2

+
x
+
1
=
0

{\displaystyle x^{2}+x+1=0}

has solutions

x
=

−
1
+

−
3

2

and

x
=

−
1
−

−
3

2

.

{\displaystyle x={\frac {-1+{\sqrt {-3}}}{2}}\quad \quad {\text{and}}\quad \quad x={\frac {-1-{\sqrt {-3}}}{2}}.}

Since

−
3

{\displaystyle {\sqrt {-3}}}

is not any real number, both of these solutions for x are complex numbers.

= Exponential and logarithmic equations

=

An exponential equation is one which has the form

a

x

=
b

{\displaystyle a^{x}=b}

for

a
>
0

{\displaystyle a>0}

, which has solution

x
=

log

a

⁡
b
=

ln
⁡
b

ln
⁡
a

{\displaystyle x=\log _{a}b={\frac {\ln b}{\ln a}}}

when

b
>
0

{\displaystyle b>0}

. Elementary algebraic techniques are used to rewrite a given equation in the above way before arriving at the solution. For example, if

3
⋅

2

x
−
1

+
1
=
10

{\displaystyle 3\cdot 2^{x-1}+1=10}

then, by subtracting 1 from both sides of the equation, and then dividing both sides by 3 we obtain

2

x
−
1

=
3

{\displaystyle 2^{x-1}=3}

whence

x
−
1
=

log

2

⁡
3

{\displaystyle x-1=\log _{2}3}

or

x
=

log

2

⁡
3
+
1.

{\displaystyle x=\log _{2}3+1.}

A logarithmic equation is an equation of the form

l
o

g

a

(
x
)
=
b

{\displaystyle log_{a}(x)=b}

for

a
>
0

{\displaystyle a>0}

, which has solution

x
=

a

b

.

{\displaystyle x=a^{b}.}

For example, if

4

log

5

⁡
(
x
−
3
)
−
2
=
6

{\displaystyle 4\log _{5}(x-3)-2=6}

then, by adding 2 to both sides of the equation, followed by dividing both sides by 4, we get

log

5

⁡
(
x
−
3
)
=
2

{\displaystyle \log _{5}(x-3)=2}

whence

x
−
3
=

5

2

=
25

{\displaystyle x-3=5^{2}=25}

from which we obtain

x
=
28.

{\displaystyle x=28.}

= Radical equations

=

A radical equation is one that includes a radical sign, which includes square roots,

x

,

{\displaystyle {\sqrt {x}},}

cube roots,

x

3

{\displaystyle {\sqrt[{3}]{x}}}

, and nth roots,

x

n

{\displaystyle {\sqrt[{n}]{x}}}

. Recall that an nth root can be rewritten in exponential format, so that

x

n

{\displaystyle {\sqrt[{n}]{x}}}

is equivalent to

x

1
n

{\displaystyle x^{\frac {1}{n}}}

. Combined with regular exponents (powers), then

x

3

2

{\displaystyle {\sqrt[{2}]{x^{3}}}}

(the square root of x cubed), can be rewritten as

x

3
2

{\displaystyle x^{\frac {3}{2}}}

. So a common form of a radical equation is

x

m

n

=
a

{\displaystyle {\sqrt[{n}]{x^{m}}}=a}

(equivalent to

x

m
n

=
a

{\displaystyle x^{\frac {m}{n}}=a}

) where m and n are integers. It has real solution(s):

For example, if:

(
x
+
5

)

2

/

3

=
4

{\displaystyle (x+5)^{2/3}=4}

then

x
+
5

=
±
(

4

)

3

,

x
+
5

=
±
8
,

x

=
−
5
±
8
,

{\displaystyle {\begin{aligned}x+5&=\pm ({\sqrt {4}})^{3},\\x+5&=\pm 8,\\x&=-5\pm 8,\end{aligned}}}

and thus

x
=
3

or

x
=
−
13

{\displaystyle x=3\quad {\text{or}}\quad x=-13}

= System of linear equations

=

There are different methods to solve a system of linear equations with two variables.

Elimination method

An example of solving a system of linear equations is by using the elimination method:

{

4
x
+
2
y

=
14

2
x
−
y

=
1.

{\displaystyle {\begin{cases}4x+2y&=14\\2x-y&=1.\end{cases}}}

Multiplying the terms in the second equation by 2:

4
x
+
2
y
=
14

{\displaystyle 4x+2y=14}

4
x
−
2
y
=
2.

{\displaystyle 4x-2y=2.}

Adding the two equations together to get:

8
x
=
16

{\displaystyle 8x=16}

which simplifies to

x
=
2.

{\displaystyle x=2.}

Since the fact that

x
=
2

{\displaystyle x=2}

is known, it is then possible to deduce that

y
=
3

{\displaystyle y=3}

by either of the original two equations (by using 2 instead of x ) The full solution to this problem is then

{

x
=
2

y
=
3.

{\displaystyle {\begin{cases}x=2\\y=3.\end{cases}}}

This is not the only way to solve this specific system; y could have been resolved before x.

Substitution method

Another way of solving the same system of linear equations is by substitution.

{

4
x
+
2
y

=
14

2
x
−
y

=
1.

{\displaystyle {\begin{cases}4x+2y&=14\\2x-y&=1.\end{cases}}}

An equivalent for y can be deduced by using one of the two equations. Using the second equation:

2
x
−
y
=
1

{\displaystyle 2x-y=1}

Subtracting

2
x

{\displaystyle 2x}

from each side of the equation:

2
x
−
2
x
−
y

=
1
−
2
x

−
y

=
1
−
2
x

{\displaystyle {\begin{aligned}2x-2x-y&=1-2x\\-y&=1-2x\end{aligned}}}

and multiplying by −1:

y
=
2
x
−
1.

{\displaystyle y=2x-1.}

Using this y value in the first equation in the original system:

4
x
+
2
(
2
x
−
1
)

=
14

4
x
+
4
x
−
2

=
14

8
x
−
2

=
14

{\displaystyle {\begin{aligned}4x+2(2x-1)&=14\\4x+4x-2&=14\\8x-2&=14\end{aligned}}}

Adding 2 on each side of the equation:

8
x
−
2
+
2

=
14
+
2

8
x

=
16

{\displaystyle {\begin{aligned}8x-2+2&=14+2\\8x&=16\end{aligned}}}

which simplifies to

x
=
2

{\displaystyle x=2}

Using this value in one of the equations, the same solution as in the previous method is obtained.

{

x
=
2

y
=
3.

{\displaystyle {\begin{cases}x=2\\y=3.\end{cases}}}

This is not the only way to solve this specific system; in this case as well, y could have been solved before x.

= Other types of systems of linear equations

Inconsistent systems

In the above example, a solution exists. However, there are also systems of equations which do not have any solution. Such a system is called inconsistent. An obvious example is

{

x
+
y

=
1

0
x
+
0
y

=
2

.

{\displaystyle {\begin{cases}{\begin{aligned}x+y&=1\\0x+0y&=2\,.\end{aligned}}\end{cases}}}

As 0≠2, the second equation in the system has no solution. Therefore, the system has no solution.
However, not all inconsistent systems are recognized at first sight. As an example, consider the system

{

4
x
+
2
y

=
12

−
2
x
−
y

=
−
4

.

{\displaystyle {\begin{cases}{\begin{aligned}4x+2y&=12\\-2x-y&=-4\,.\end{aligned}}\end{cases}}}

Multiplying by 2 both sides of the second equation, and adding it to the first one results in

0
x
+
0
y
=
4

,

{\displaystyle 0x+0y=4\,,}

which clearly has no solution.

Undetermined systems

There are also systems which have infinitely many solutions, in contrast to a system with a unique solution (meaning, a unique pair of values for x and y) For example:

{

4
x
+
2
y

=
12

−
2
x
−
y

=
−
6

{\displaystyle {\begin{cases}{\begin{aligned}4x+2y&=12\\-2x-y&=-6\end{aligned}}\end{cases}}}

Isolating y in the second equation:

y
=
−
2
x
+
6

{\displaystyle y=-2x+6}

And using this value in the first equation in the system:

4
x
+
2
(
−
2
x
+
6
)
=
12

4
x
−
4
x
+
12
=
12

12
=
12

{\displaystyle {\begin{aligned}4x+2(-2x+6)=12\\4x-4x+12=12\\12=12\end{aligned}}}

The equality is true, but it does not provide a value for x. Indeed, one can easily verify (by just filling in some values of x) that for any x there is a solution as long as

y
=
−
2
x
+
6

{\displaystyle y=-2x+6}

. There is an infinite number of solutions for this system.

Over- and underdetermined systems

Systems with more variables than the number of linear equations are called underdetermined. Such a system, if it has any solutions, does not have a unique one but rather an infinitude of them. An example of such a system is

{

x
+
2
y

=
10

y
−
z

=
2.

{\displaystyle {\begin{cases}{\begin{aligned}x+2y&=10\\y-z&=2.\end{aligned}}\end{cases}}}

When trying to solve it, one is led to express some variables as functions of the other ones if any solutions exist, but cannot express all solutions numerically because there are an infinite number of them if there are any.
A system with a higher number of equations than variables is called overdetermined. If an overdetermined system has any solutions, necessarily some equations are linear combinations of the others.

References

Leonhard Euler, Elements of Algebra, 1770. English translation Tarquin Press, 2007, ISBN 978-1-899618-79-8, also online digitized editions 2006, 1822.
Charles Smith, A Treatise on Algebra, in Cornell University Library Historical Math Monographs.
Redden, John. Elementary Algebra Archived 2016-06-10 at the Wayback Machine. Flat World Knowledge, 2011