Conditionals, contradictions, and identities

There may be many ways to separate equations into distinct categories but in this chapter, we discuss a particular categorization that is often helpful.

We can identify equations that have no solution. That is, no number exists that would make the equation a true statement if substituted for the unknown quantity. We will call this type of equation inconsistent or contradictory.

Another type of equation remains a true statement when any number whatsoever is put in place of the unknown quantity. This type, we will call an identity.

Finally, there are equations that have exactly one solution or possibly some countable number of solutions. We will call these conditional equations.

inconsistent equations

Example 1

Consider the statement $2x+1=2x+3$2x+1=2x+3.

No matter what number we choose to try for $x$x, we will end up with an obviously false statement like $1=3$1=3 or $1=0$1=0. For example, let $x$x be a particular number $a$a. Then

$2a+1$2a+1	$=$=	$2a+3$2a+3
$2a$2a	$=$=	$2a+2$2a+2   We subtracted $1$1 from both sides.
$a$a	$=$=	$a+1$a+1   We divided both sides by $2$2.
$0$0	$=$=	$1$1   We subtracted the number $a$a from both sides.

This is a contradiction. Therefore, it could not have been true that $x$x was any particular number $a$a.

 

Example 2

When solving a system of simultaneous equations, a pair of statements like the following can sometimes arise.

$x+2y=11$x+2y=11
$3x+6y=28$3x+6y=28

If the first equation is multiplied by $3$3, we get the equivalent statement

$3x+6y=33$3x+6y=33

and there is no combination of numbers $x$x and $y$y  that can make $3x+6y$3x+6y be equal to both $33$33 and $28$28. We say this pair of equations is contradictory or inconsistent.

 

identities

Example 3

The statement $5x-3x=2x$5x−3x=2x is an identity. It is true no matter what the value of $x$x. 

We use identities similar to this one with hardly a thought whenever we simplify an algebraic expression. We often use an 'equals' sign with three lines $\equiv$≡ to indicate an identity.

Here is another example: $\frac{y+1}{3}+\frac{y-1}{4}\equiv\frac{7y+1}{12}$y+13​+y−14​≡7y+112​. We found this identity by adding the fractions on the left in the usual way. You should check that it is true for various randomly chosen values of $y$y.

 

Example 4

The statement defining the tangent function in trigonometry, $\tan x=\frac{\sin x}{\cos x}$tanx=sinxcosx​ is an identity. We cannot solve it to find a particular value for $x$x. It is a definition and is therefore true for all values of $x$x (except values of $x$x that make $\cos x=0$cosx=0).

 

conditional equations

Example 5

The equation $3x+5x=9x$3x+5x=9x may look, at first glance, like one that cannot be true for any number $x$x. We might think this because if both sides are divided by $x$x, which we can do if $x\ne0$x≠0, then the equation says $3+5=9$3+5=9  and this is false. 

However, if we gather the like terms by subtracting $9x$9x from both sides, we obtain an equivalent statement:

$3x+5x-9x=0$3x+5x−9x=0

and then, taking the common factor $x$x outside a bracket we have

$x(3+5-9)=0$x(3+5−9)=0

For this statement to be true, we must have either $x=0$x=0 or $3+5-9=0$3+5−9=0. So, $x=0$x=0 is the solution we are looking for.

 

We speak of the solution set for an equation. This is the empty set $\varnothing$∅ if there are no solutions or it could be the interval $(-\infty,\infty)$(−∞,∞) or some other collection of intervals in the case of an identity, or it could be a single number $\left\{a\right\}${a} or a few numbers, $\left\{a,b,c\right\}${a,b,c} in the case of a conditional equation.

Example 6

What type of equation is $\frac{x+3}{5}-\frac{x-5}{3}=1$x+35​−x−53​=1? Find its solution set.

If we make the substitution $x=0$x=0, we obtain $0=1$0=1, which is false. So, the equation cannot be an identity.

We try adding the fractions on the left. (In effect, we are creating a chain of identities.)

$\frac{x+3}{5}-\frac{x-5}{3}$x+35​−x−53​	$=$=	$\frac{3x+9}{15}+\frac{5x-25}{15}$3x+915​+5x−2515​
	$=$=	$\frac{8x-16}{15}$8x−1615​

But, this has to be equal to the fixed right-hand side. This is a condition and it will result in the equation having just one solution.

$\frac{8x-16}{15}$8x−1615​	$=$=	$1$1
$\therefore\ \ 8x-16$∴  8x−16	$=$=	$15$15
$8x$8x	$=$=	$31$31
$x$x	$=$=	$\frac{31}{8}$318​

The solution set is the single solution $\left\{3\frac{7}{8}\right\}${378​} and so, the equation was a conditional equation.

 

Worked Examples

Question 1

Question 2

Question 3