Equations

Lesson

The 'equals' sign $=$= is used to indicate that whatever is on the left of the sign is the same as whatever is on the right.

Its use was introduced by Robert Recorde in 1557. He explained the use of the two parallel lines by saying '... no two things can be more equal'.

When there are algebraic expressions at the two ends of the equals sign, typically with unspecified quantities $x$`x` and $y$`y`, it may be that a restricted number of $x,y$`x`,`y` pairs exists such that the statement is true. For example, the statement $2x=3y$2`x`=3`y` is not true for every pair of numbers $x$`x` and $y$`y`, but it is true for special pairs like $(6,4)$(6,4), $(9,6)$(9,6), $(11,7\frac{1}{3})$(11,713) and so on.

The pairs of numbers for which the statement is true are said to be *solutions *and the statement itself is called an *equation*.

An equation can have no solutions, exactly one solution or many solutions. There should also be non-solutions.

If it should happen that a statement is true for *every *possible number or combination of numbers that can be substituted for the unspecified variables, then we call the statement an *identity*. The sign used for an identity has three parallel lines rather than two, like this: $\equiv$≡. It means that the expressions at the two ends of the sign are equivalent. For example, we could write $\frac{2x}{5}\equiv\frac{4x^2}{10x}$2`x`5≡4`x`210`x`.

An identity cannot be 'solved' in the way an equation can because it is true for all values of the variable - which is not very helpful if one is looking for a particular solution.

Identities are very useful when it is desirable to express an algebraic statement in an alternative but equally true form.

The equation $14x^2+x-3=0$14`x`2+`x`−3=0 has two solutions. That is, there are two values of $x$`x` that make the statement true. However, it is not easy to find the solutions until it is realised that the left-hand side of the equation can be factorised. In doing this we are making use of the identity

$14x^2+x-3\equiv(7x-3)(2x+1)$14`x`2+`x`−3≡(7`x`−3)(2`x`+1)

Since these two expressions are the same for all values of $x$`x`, it must be true that $(7x-3)(2x+1)=0$(7`x`−3)(2`x`+1)=0 is a valid statement with the same solutions as the original equation. In this form the equation is easy to solve.

In trigonometry, much use is made of what is called the Pythagorean identity. We write

$\sin^2\theta+\cos^2\theta\equiv1$`s``i``n`2`θ`+`c``o``s`2`θ`≡1

It is derived from the theorem about the relation between the sides of a right-angled triangle, $a^2+b^2=c^2$`a`2+`b`2=`c`2. *This *statement *is *an equation, however, because a random choice of numbers $a,b,c$`a`,`b`,`c` would almost certainly not make the statement true. Thus, the statement has non-solutions.

In trigonometry, we define the tangent function for all values of $x$`x` such that $\cos x\ne0$`c``o``s``x`≠0 by the identity $\tan x\equiv\frac{\sin x}{\cos x}$`t``a``n``x`≡`s``i``n``x``c``o``s``x`.

The statement $\tan x\equiv\frac{\sin x}{\cos x}$`t``a``n``x`≡`s``i``n``x``c``o``s``x` is true for all $x$`x` in the domain of $\tan x$`t``a``n``x` because that is how the tangent function is defined.

When there is doubt about whether a statement is an identity, it is not sufficient to show that for every choice of a solution that we have tried the two sides of the statement are the same. This strategy fails because it is not possible to check every candidate solution one-by-one. We might miss one!

Either, we have to show that one side of the proposed identity can be transformed into the other by a series of logical algebraic manipulations, or, in an extreme case, we might have to show that the contrary assumption, that the statement is not an identity, leads to a logical absurdity.

For example, the statement $\frac{2x}{5}\equiv\frac{4x^2}{10x}$2`x`5≡4`x`210`x` is certainly an identity (when $x\ne0$`x`≠0) because we can obtain the right-hand side from the left by multiplying by $1$1 in the form $\frac{2x}{2x}$2`x`2`x`.

Another example concerns what are called the hyperbolic sine and cosine functions. They have special symbols but, for simplicity, we will call them $g(x)$`g`(`x`) and $f(x)$`f`(`x`) with the definitions

$g(x)=\frac{1}{2}\left(e^x+e^{-x}\right)$`g`(`x`)=12(`e``x`+`e`−`x`) and

$f(x)=\frac{1}{2}\left(e^x-e^{-x}\right)$`f`(`x`)=12(`e``x`−`e`−`x`)

For these there is an identity $\left[g(x)\right]^2-\left[f(x)\right]^2\equiv1$[`g`(`x`)]2−[`f`(`x`)]2≡1, which is similar to the Pythagorean identity in trigonometry.

The proof strategy will be to transform the left-hand side of the statement into the right. We wish to simplify $\left[g(x)\right]^2-\left[f(x)\right]^2$[`g`(`x`)]2−[`f`(`x`)]2.

We factorise by the difference of two squares to obtain the equivalent expression $\left(g(x)+f(x)\right)\left(g(x)-f(x)\right)$(`g`(`x`)+`f`(`x`))(`g`(`x`)−`f`(`x`)).

From the definitions of $g(x)$`g`(`x`) and $f(x)$`f`(`x`), we see that this expression is equivalent to $e^x.e^{-x}$`e``x`.`e`−`x` which, in turn, is equivalent to $e^0=1$`e`0=1.

The reasoning did not depend on any particular values of $x$`x`. So, the statement must be true for all $x$`x`.