Editing Elementary algebra (section)

== Solving algebraic equations ==
{{see also|Equation solving}}
[[File:Algebraproblem.jpg|thumb|A typical algebra problem.]]

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

=== Linear equations with one variable ===
{{Main|Linear equation}}

Linear equations are so-called, because when they are plotted, they describe a straight line. The simplest equations to solve are [[linear equation]]s 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: <math>2x + 4 = 12</math> where  {{mvar|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.<ref>{{Cite book|title=All the Math You'll Ever Need|author=Slavin, Steve|publisher=[[John Wiley & Sons]]|year=1989|isbn=0-471-50636-2|page=[https://archive.org/details/allmathyoullever00slav/page/72 72]|url=https://archive.org/details/allmathyoullever00slav/page/72}}</ref> This problem and its solution are as follows:
[[File:Divide large.gif|alt=|thumb|Solving for x]]
{|
|-
| 1. Equation to solve:
| <math>2x + 4 = 12</math>
|-
| 2. Subtract 4 from both sides:
| <math>2x + 4 - 4 = 12 - 4</math>
|-
| 3. This simplifies to:
| <math>2x = 8</math>
|-
| 4. Divide both sides by 2:
| <math>\frac{2x}{2} = \frac{8}{2}</math>
|-
| 5. This simplifies to the solution:
| <math>x = 4</math>
|}
In words: the child is 4 years old.

The general form of a linear equation with one variable, can be written as: <math>ax+b=c</math>

Following the same procedure (i.e. subtract {{mvar|b}} from both sides, and then divide by {{mvar|a}}), the general solution is given by <math>x=\frac{c-b}{a}</math>

=== Linear equations with two variables ===
[[File:Linear-equations-two-unknowns.svg|thumb|right|Solving two linear equations with a unique solution at the point that they intersect.]]
A linear equation with two variables has many (i.e. an infinite number of) solutions.<ref>Sinha, ''The Pearson Guide to Quantitative Aptitude for CAT 2/e''Publisher: Pearson Education India, 2010, {{ISBN|8131723666}}, 9788131723661, 599 pages, [https://books.google.com/books?id=eOnaFSKRSR0C&q=many+solutions&pg=PA195 page 195]</ref> For example:

:Problem in words: A father is 22 years older than his son. How old are they?
:Equivalent equation: <math>y = x + 22</math> where {{mvar|y}} is the father's age, {{mvar|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
: <math>\begin{align}
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{align}</math>

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):<ref>[[Cynthia Y. Young]], ''Precalculus'', Publisher John Wiley & Sons, 2010, {{ISBN|0471756849}}, 9780471756842, 1175 pages, [https://books.google.com/books?id=9HRLAn326zEC&dq=linear+equation++two+variables++many+solutions&pg=PA699 page 699]</ref>
:<math>\begin{cases}
y = x + 22 & \text{First equation}\\
y = 2x + 10 & \text{Second equation}
\end{cases}</math>
:<math>\begin{align}
&&&\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{align}</math>

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|System of linear equations]]'''.

=== Quadratic equations ===
{{Main|Quadratic equation}}
[[File:Quadratic-equation.svg|thumb|right|Quadratic equation plot of <math>y = x^2 + 3x - 10</math> showing its roots at <math>x = -5</math> and <math>x = 2</math>, and that the quadratic can be rewritten as <math>y = (x + 5)(x - 2)</math>
]]
A quadratic equation is one which includes a term with an exponent of 2, for example, <math>x^2</math>,<ref>Mary Jane Sterling, ''Algebra II For Dummies'', Publisher: John Wiley & Sons, 2006, {{ISBN|0471775819}}, 9780471775812, 384 pages, [https://books.google.com/books?id=_0rTMuSpTY0C&dq=quadratic+equations&pg=PA37 page 37]</ref> and no term with higher exponent. The name derives from the Latin ''quadrus'', meaning square.<ref>John T. Irwin, ''The Mystery to a Solution: Poe, Borges, and the Analytic Detective Story'', Publisher JHU Press, 1996, {{ISBN|0801854660}}, 9780801854668, 512 pages, [https://books.google.com/books?id=jsxTenuOQKgC&dq=quadratic+quadrus&pg=PA372 page 372]</ref> In general, a quadratic equation can be expressed in the form <math>ax^2 + bx + c = 0</math>,<ref>Sharma/khattar, ''The Pearson Guide To Objective Mathematics For Engineering Entrance Examinations, 3/E'', Publisher Pearson Education India, 2010, {{ISBN|8131723631}}, 9788131723630, 1248 pages, [https://books.google.com/books?id=2v-f9x7-FlsC&dq=quadratic%20equations%20%20ax2%20%2B%20bx%20%2B%20c%20%3D%200&pg=RA13-PA33 page 621]</ref> where {{mvar|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 <math>ax^2</math>, which is known as the quadratic term. Hence <math>a \neq 0</math>, and so we may divide by {{mvar|a}} and rearrange the equation into the standard form

: <math>x^2 + px + q = 0 </math>

where <math>p = \frac{b}{a}</math> and <math>q = \frac{c}{a}</math>. Solving this, by a process known as [[completing the square]], leads to the [[quadratic formula]]

:<math>x=\frac{-b \pm \sqrt {b^2-4ac}}{2a},</math>

where [[plus–minus sign|the symbol "±"]] indicates that both

: <math> x=\frac{-b + \sqrt {b^2-4ac}}{2a}\quad\text{and}\quad x=\frac{-b - \sqrt {b^2-4ac}}{2a}</math>

are solutions of the quadratic equation.

Quadratic equations can also be solved using [[factorization]] (the reverse process of which is [[polynomial expansion|expansion]], but for two [[linear function|linear terms]] is sometimes denoted [[FOIL rule|foiling]]).  As an example of factoring:

: <math>x^{2} + 3x - 10 = 0, </math>

which is the same thing as

: <math>(x + 5)(x - 2) = 0. </math>

It follows from the [[zero-product property]] that either <math>x = 2</math> or <math>x = -5</math> 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,

: <math>x^{2} + 1 = 0 </math>

has no real number solution since no real number squared equals −1.
Sometimes a quadratic equation has a root of [[multiplicity (mathematics)|multiplicity]] 2, such as:

: <math>(x + 1)^2 = 0. </math>

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

:<math>[x-(-1)][x-(-1)]=0.</math>

====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 number]]s, [[imaginary number]]s, 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

:<math>x^2+x+1=0</math>

has solutions

:<math>x=\frac{-1 + \sqrt{-3}}{2}  \quad \quad \text{and} \quad \quad x=\frac{-1-\sqrt{-3}}{2}.</math>

Since <math>\sqrt{-3}</math> is not any real number, both of these solutions for ''x'' are complex numbers.

=== Exponential and logarithmic equations ===
{{Main|Logarithm}}
[[File:Binary logarithm plot with ticks.svg|right|thumb|upright=1.35|alt=Graph showing a logarithm curves, which crosses the ''x''-axis where ''x'' is 1 and extend towards minus infinity along the ''y''-axis.|The [[graph of a function|graph]] of the logarithm to base 2 crosses the [[x axis|''x'' axis]] (horizontal axis) at 1 and passes through the points with [[coordinate]]s {{nowrap|(2, 1)}}, {{nowrap|(4, 2)}}, and {{nowrap|(8, 3)}}. For example, {{nowrap|log<sub>2</sub>(8) {{=}} 3}}, because {{nowrap|2<sup>3</sup> {{=}} 8.}} The graph gets arbitrarily close to the ''y'' axis, but [[asymptotic|does not meet or intersect it]].]]

An exponential equation is one which has the form <math>a^x = b</math> for <math>a > 0</math>,<ref>Aven Choo, ''LMAN OL Additional Maths Revision Guide 3'', Publisher Pearson Education South Asia, 2007, {{ISBN|9810600011}}, 9789810600013, [https://books.google.com/books?id=NsBXDMrzcJIC&dq=%22+exponential+equation+%22+aX+%3D+b&pg=RA2-PA29 page 105]</ref> which has solution

: <math>x = \log_a b = \frac{\ln b}{\ln a}</math>

when <math>b > 0</math>. Elementary algebraic techniques are used to rewrite a given equation in the above way before arriving at the solution. For example, if

: <math>3 \cdot 2^{x - 1} + 1 = 10</math>

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

: <math>2^{x - 1} = 3</math>

whence

: <math>x - 1 = \log_2 3</math>

or

: <math>x = \log_2 3 + 1.</math>

A logarithmic equation is an equation of the form <math>log_a(x) = b</math> for <math>a > 0</math>, which has solution

: <math>x = a^b.</math>

For example, if

: <math>4\log_5(x - 3) - 2 = 6</math>

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

: <math>\log_5(x - 3) = 2</math>

whence

: <math>x - 3 = 5^2 = 25</math>

from which we obtain

: <math>x = 28.</math>

=== Radical equations ===
{{Image frame|align=right|width=150|caption=Radical equation showing two ways to represent the same expression. The triple bar means the equation is true for all values of ''x''|content=<math>\overset{}{\underset{}{\sqrt[2]{x^3} \equiv x^{\frac 3 2} } }</math>}}
A radical equation is one that includes a radical sign, which includes [[square root]]s, <math>\sqrt{x},</math>  [[cube root]]s, <math>\sqrt[3]{x}</math>, and [[nth root|''n''th roots]], <math>\sqrt[n]{x}</math>. Recall that an ''n''th root can be rewritten in exponential format, so that <math>\sqrt[n]{x}</math> is equivalent to <math>x^{\frac{1}{n}}</math>. Combined with regular exponents (powers), then  <math>\sqrt[2]{x^3}</math> (the square root of {{mvar|x}} cubed), can be rewritten as <math>x^{\frac{3}{2}}</math>.<ref>John C. Peterson, ''Technical Mathematics With Calculus'', Publisher Cengage Learning, 2003, {{ISBN|0766861899}}, 9780766861893, 1613 pages, [https://books.google.com/books?id=PGuSDjHvircC&dq=%22+radical+equation%22&pg=PA525 page 525]</ref> So a common form of a radical equation is <math> \sqrt[n]{x^m}=a</math> (equivalent to <math> x^\frac{m}{n}=a</math>) where {{mvar|m}} and {{mvar|n}} are [[integer]]s. It has [[real number|real]] solution(s):
{| class="wikitable" style="text-align:center"
|- style="vertical-align:top"
!{{mvar|n}} is odd
!{{mvar|n}} is even<br />and <math>a \ge 0</math>
!{{mvar|n}} '''and''' {{mvar|m}} are '''even'''<br />'''and''' <math>a<0</math>
!{{mvar|n}} '''is even''', {{mvar|m}} '''is odd''', '''and''' <math>a<0</math>
|-
|<math>x = \sqrt[n]{a^m}</math>

equivalently

:<math>x = \left(\sqrt[n]a\right)^m</math>
|<math>x = \pm \sqrt[n]{a^m}</math>

equivalently

:<math>x = \pm \left(\sqrt[n]a\right)^m</math>

|<math>x=\pm \sqrt[n]{a^m}</math>
|no real solution
|}

For example, if:

:<math>(x + 5)^{2/3} = 4</math>

then

: <math>\begin{align}
x + 5 & = \pm (\sqrt{4})^3,\\
x + 5 & = \pm 8,\\
x &  = -5 \pm 8,
\end{align}</math>
and thus 
:<math>x = 3 \quad \text{or}\quad x = -13</math>

=== System of linear equations ===
{{Main|System of linear equations}}

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

==== Elimination method ====
[[File:Intersecting Lines.svg|thumb|right|The solution set for the equations <math>x - y = -1</math> and <math>3x + y = 9</math> is the single point (2,&nbsp;3).]]

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

: <math>\begin{cases}4x + 2y&= 14 \\
2x - y&= 1.\end{cases} </math>

Multiplying the terms in the second equation by 2:

: <math>4x + 2y = 14 </math>
: <math>4x - 2y = 2. </math>

Adding the two equations together to get:

: <math>8x = 16 </math>

which simplifies to

: <math>x = 2. </math>

Since the fact that <math>x = 2</math> is known, it is then possible to deduce that <math>y = 3</math>  by either of the original two equations (by using ''2'' instead of {{mvar|x}} ) The full solution to this problem is then

: <math>\begin{cases} x = 2 \\ y = 3. \end{cases}</math>

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

==== Substitution method ====

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

: <math>\begin{cases}4x + 2y &= 14
\\ 2x - y &= 1.\end{cases} </math>

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

: <math>2x - y = 1 </math>

Subtracting <math>2x</math> from each side of the equation:

: <math>\begin{align}2x - 2x - y & = 1 - 2x  \\
- y & = 1 - 2x
\end{align}</math>

and multiplying by −1:

: <math> y = 2x - 1. </math>

Using this {{mvar|y}} value in the first equation in the original system:

: <math>\begin{align}4x + 2(2x - 1) &= 14\\
4x + 4x - 2 &= 14 \\
8x - 2 &= 14 \end{align}</math>

Adding ''2'' on each side of the equation:

: <math>\begin{align}8x - 2 + 2 &= 14 + 2 \\
8x &= 16 \end{align}</math>

which simplifies to

: <math>x = 2 </math>

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

: <math>\begin{cases} x = 2 \\ y = 3. \end{cases}</math>

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

=== Other types of systems of linear equations ===

==== Inconsistent systems ====
[[File:Parallel Lines.svg|thumb|right|The equations <math>3x + 2y = 6</math> and <math>3x + 2y = 12</math> are parallel and cannot intersect, and is unsolvable.]]
[[File:Quadratic-linear-equations.svg|thumb|right|Plot of a quadratic equation (red) and a linear equation (blue) that do not intersect, and consequently for which there is no common solution.]]

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 system|inconsistent]]. An obvious example is

: <math>\begin{cases}\begin{align} x + y &= 1 \\
0x + 0y &= 2\,. \end{align} \end{cases}</math>

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 
: <math>\begin{cases}\begin{align}4x + 2y &= 12 \\
-2x - y &= -4\,. \end{align}\end{cases}</math>

Multiplying by 2 both sides of the second equation, and adding it to the first one results in
: <math>0x+0y = 4 \,,</math>
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 {{mvar|x}} and {{mvar|y}}) For example:

: <math>\begin{cases}\begin{align}4x + 2y & = 12 \\
-2x - y & = -6 \end{align}\end{cases}</math>

Isolating {{mvar|y}} in the second equation:

: <math>y = -2x + 6 </math>

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

: <math>\begin{align}4x + 2(-2x + 6) = 12 \\
4x - 4x + 12 = 12 \\
12 = 12 \end{align}</math>

The equality is true, but it does not provide a value for {{mvar|x}}. Indeed, one can easily verify (by just filling in some values of {{mvar|x}}) that for any {{mvar|x}} there is a solution as long as <math>y = -2x + 6</math>. 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 system|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

: <math>\begin{cases}\begin{align}x + 2y & = 10\\
y - z  & = 2 .\end{align}\end{cases}</math>

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 [[Number|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 system|overdetermined]]. If an overdetermined system has any solutions, necessarily some equations are [[linear combination]]s of the others.