\chapter{Algebra} 
   
\section{Introduction}   
   
Algebra \pron{£h£al£d£ibr£e}, like \nuvo{arithmetic} \pron{£e£hr£i£fmet£ik}, deals with   
numbers.\index{algebra} Algebra is an extension of arithmetic.  Both subjects employ the fundamental   
operations. In arithmetic you use numbers whose values are known; you operate with these and obtain   
definite numerical \pron{nj£U£hmer£ikl} results. Whereas in algebra, while you may use definite numbers   
on accasions, you are, in the main, concerned with general expressions and general results, in which   
letters or other symbols represent numbers not named or specified.   
   
There is a rule to determine the area of a rectangle which might be expressed in the form:   
`\emph{the area of a rectangle in square meters is equal to the product of the length in metres by   
  the breadth in metres}'. This rule is shortened by employing letters as symbols, to represent the   
quantities. Thus:   
   
Let $l$ represent the `lenght in metres', $b$ represent the `breadth in metres' and $A$ represent `the area   
in square metres'. With those symbols you can now write the above rule in the form: $A = l\times b$.    
   
This shortened form shows the rule for finding the area of any rectangle; it is a general rule, and   
is called a \nuvo{formula} \pron{£hf£Omj£ul£e}. $A$ is said to be the \nuvo{subject} of the formula.  In   
that formula $A$ is expressed in terms of $l$ and $b$.\index{term!in terms of} You can change the   
subject of the formula, for example $b=A/l$.\index{subject!change}   
   
If you wish to find the numerical value of an algebraic expression for definite numerical values of   
the letters in it, you should \nuvo{substitute} \pron{£hs£vbst£itj£Ut} the numerical values for the   
letters.  So, when $l=3$ and $b=2$, substituting the given values leads to $A=6$.   
   
It frequently happens that an expression, or part of an expression, is to be operated as a whole.   
For example, suppose that you wish to write in algebraic \pron{£b£al£d£i£hbr£q£ik} symbols `Twice the sum   
of $a$ and $b$'.  Brackets are used to enclose the part which is to be operated on as a whole ---   
namely, $a+b$. You write $2\parent{a+b}$.   
   
The expression $3a$ denotes a multiple of $a$ and the number $3$, which indicates the multiple, is   
called the \nuvo{coefficient} \pron{£bk£w£i£hf£i£snt} of $a$. In $3ab$, $3$ is the coefficient of $ab$,   
$3a$ is the coefficient of $b$ and $3b$ is the coefficient of $a$.   
   
In a expression, terms which involve the same letter, and differ only in the coefficients of this   
letter, are called \emph{like terms}.\index{term!like} Thus in the expression $3a+5b-2a+4b$, $3a$   
and $2a$ are like terms.   
   
\section{Equations}   
   
An \nuvo{equation} \pron{£i£hkw£q£jn} is an expression that represents the \nuvo{equality}   
\pron{£i£hkw£ol£et£i} of two expressions involving constants \pron{£hk£onst(£e)nt} or variables \pron{£hv£Pr£Qbl}.   
For example, the equation $A= §pr^2$ \emph{equates} \pron{£i£hkw£qt} the area $A$ of a circle \emph{to}   
the product $§pr^2$.\index{equate}   
   
Let $x$ be an \nuvo{unknown} number. To \nuvo{solve} \pron{s£olv} the equation $5x=40$ is to find the   
value(s) of $x$ which satisfy(fies) the equality. Those values are called the solutions   
\pron{s£e£hl£U£sn} of the equation.\index{solution}   
   
\begin{prop}   
\begin{itemize}   
\item If the same number is added to, or subtracted from, both sides of an equation, the two sides   
  will again be equal.   
\item If both sides of an equation are multiplied or divided by the same number, the two sides of   
  the new equation will be equal.   
\end{itemize}   
\end{prop}   
   
\subsection{Example}   
Solve the equation: $6x-5=2x+9$.   
   
The general plan adopted is to collect the terms involving the unknown on the left (hand) side, and   
the other terms on the right. Transferring the $x$ terms from the right side you get:   
$$6x-2x-5=9.$$   
Transferring the $-5$,   
\begin{align*}   
6x-2x&=9+5\\   
\wasytherefore\qquad 4x&=14\\   
\text{and}\qquad x&=\dfrac{7}{2}.   
\end{align*}   
The solution is \Dfrac{7}{2}. The \emph{solution set}\index{solution!set} is \accol{\dfrac{7}{2}}   
\idest\  the set containing \Frac{7}{2} as its only element.   
   
\begin{rem}   
The sign $\wasytherefore$ denotes `therefore'.\index{$\wasytherefore$\quad}   
\end{rem}   
   
\section{Simultaneous equations}   
\emph{Simultaneous} \pron{£bs£im£el£ht£qn£i£es} \emph{equations} are where we have a pair (or more) of   
equations that both (or all) need solving at the same time --- hence simultaneous. There are two   
main methods: the \nuvo{elimination} \pron{£i£bl£im£i£hn£q£sn} method and the \nuvo{substitution}   
\pron{£bs£vbst£i£htj£U£sn} method.\index{equation!simultaneous}   
   
\subsection{Elimination method}   
The technique is initially to eliminate one variable to find a solution to the other one, and then   
substitute the found variable to complete the solution.   
   
\paragraph{Example} Solve the simultaneous equations   
\begin{align*}   
4x+y&=14  & (1)\\ 2x+y&=8 &(2)   
\end{align*}   
   
To eliminate one of the variables, subtract the equations, botton from top, which will give $2x=6$,   
so $x=3$.   
   
Now, we substitute this answer in the simplest equation, (2) above, to give $6+y=8$, so $y=2$.   
Therefore the solution of the simultaneous equations is $x=3$ and $y=2$. The solution set is   
\accol{\couple{3}{2}}.   
   
Note that you should always check the solution and see that it works with the other equation.   
   
\subsection{Substitution method}   
   
This is an alternative method for solving simultaneous equations.   
   
\paragraph{Example} Solve the simultaneous equations   
\begin{align*}   
4x-2y&=11  & (1)\\ 3x+y&=12 &(2)   
\end{align*}   
   
Then from equation (2), $y=12-3x$. Now substitute this into equation (1) to give   
\begin{align*}   
4x-2\parent{12-3x}&=11\\   
4x-24+6x&=11\\   
10x&=35\\   
x&=3.5   
\end{align*}   
and we are where we arrived at before. We now need to substitute $x=3.5$ into one of the equations   
to complete the solution. So $y=12-3\times 3.5 = 12-10.5 = 1.5$ and the solution set is   
\accol{\couple{3.5}{2.5}}.   
   
\section{Inequalities}   
   
\emph{Inequations} are often referred to as \emph{inequalities} also. They are solved as equations,   
except that when you have to multiply or divide \emph{both} sides by a negative number the   
inequality sign turns round.  For example, if we have $-2<5$, then $2>-5$.   
   
\paragraph{Example} Find the range of value for which $5x+3>6(x-2)$.   
   
You have to expand the brackets first to give $5x+3>6x-12$. Then transferring the $x$ term from the   
right hand side and the $3$ from the left hand side, you obtain $5x-6x > -12-3$ and so $-x>-15$. You   
have to multiply both sides by $-1$ and so $x < 15$. The solution set is the interval   
\interouv{\moinsinf}{15}.   
\endinput   
   
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