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New Zealand
Level 7 - NCEA Level 2

Defineable rational expressions


Definable Denominators

A rational expression is simply an algebraic expression that is a fraction, with polynomials in the numerator and denominator. Here are some examples of rational expressions.

$\frac{1}{x-9}$1x9 $\frac{z^2+7z-4}{z+6}$z2+7z4z+6 $\frac{7}{\left(m^2+1\right)\left(m-3\right)}$7(m2+1)(m3) $\frac{\left(p+1\right)\left(p+2\right)\left(p+3\right)}{p^2-4}$(p+1)(p+2)(p+3)p24

There is an important difference between these rational expressions and the expressions we're used to.

If we have a normal algebraic expression such as $x^2-3$x23 or $19y-2z$19y2z we can substitute in any value we like, whether it's positive or negative, large or small, and you will always be able to evaluate the expression. We say these expressions are definable for every substitution.

So what does it mean for an expression to not be definable for a substitution?

Think about the rational expression $\frac{1}{x}$1x. We can substitute any positive value of $x$x we like, to get results like $\frac{1}{2}$12, $\frac{1}{15}$115, $\frac{1}{1000}$11000 or $\frac{1}{10000012}$110000012. We could also substitute any negative value we like, to get results like $\frac{1}{-3}=-\frac{1}{3}$13=13 or $\frac{1}{-504}=-\frac{1}{504}$1504=1504.

But what about $x=0$x=0? This means we get $\frac{1}{0}$10, and we can't divide by zero! The expression breaks because dividing by zero is undefined in mathematics.

In other words, the expression $\frac{1}{x}$1x is not definable for $x=0$x=0. It is only definable for $x>0$x>0 or $x<0$x<0.

What about $\frac{1}{x-6}$1x6? What substitution isn't allowable in this expression? 

Well, as before, we can't have zero occurring in the denominator. In other words, we can't have $x-6=0$x6=0. So what value of $x$x can't we have? Solving $x-6=0$x6=0 tells us that we can't have $x=6$x=6.

This makes sense. If we did substitute in $x=6$x=6, the expression would become $\frac{1}{6-6}$166 which equals $\frac{1}{0}$10 and this is undefined.


Rational expressions are only definable for substitutions where the denominator isn't equal to zero.

What about the rational expression $\frac{m+7}{m-5}$m+7m5? For the expression to be definable we can't have the denominator $m-5$m5 equal to zero, so $m=5$m=5 isn't allowable.

If we substitute $m=5$m=5 into the expression we would get the undefined fraction $\frac{5+7}{5-5}=\frac{12}{0}$5+755=120. Notice that even though it's perfectly fine to substitute $m=5$m=5 into the numerator, the denominator forbids this substitution. 

Denominators Inside Denominators

What happens when we have the rational expression $\frac{7}{\frac{1}{p}-8}$71p8? What substitutions aren't definable here?

Denominators can't be equal to zero no matter where they are in an expression. In this expression we have two denominators to deal with, the $p$p in $\frac{1}{p}$1p, and the $\frac{1}{p}-8$1p8 in $\frac{7}{\frac{1}{p}-8}$71p8. One is actually inside the other!

$\frac{7}{\frac{1}{\left(p\right)}-8}$71(p)8 $\frac{7}{\left(\frac{1}{p}-8\right)}$7(1p8)

Because of the first denominator, we can't have $p=0$p=0. If we did, we would get $\frac{7}{\frac{1}{0}-8}$7108, and even though everything else seems fine, even one undefinable fraction breaks the whole expression!

Then, because of the second denominator, we can't have $\frac{1}{p}-8=0$1p8=0. We can solve this as follows.

$\frac{1}{p}-8$1p8 $=$= $0$0  
$\frac{1}{p}$1p $=$= $8$8 Add $8$8 to both sides
$p$p $=$= $\frac{1}{8}$18 Take the reciprocal of both sides

So $p=\frac{1}{8}$p=18 is also not an allowable substitution, along with $p=0$p=0.


Look for all denominators in a rational expression. None of them can ever be equal to zero, or else the expression is undefined.

Worked Examples

Question 1

Consider the expression $\frac{18}{\left(x+5\right)\left(x+2\right)}+\frac{5}{x^2+3x+2}$18(x+5)(x+2)+5x2+3x+2.

Which values of $x$x will cause a denominator to equal zero?

Think: Just looking at the first term, we have the rational expression $\frac{18}{\left(x+5\right)\left(x+2\right)}$18(x+5)(x+2). We can't have the denominator $\left(x+5\right)\left(x+2\right)=0$(x+5)(x+2)=0. Remember that the product of two factors will equal zero if one, the other, or both are equal to zero. In other words, either $x+5=0$x+5=0 or $x+2=0$x+2=0. Solving these tells us that we cannot substitute $x=-5$x=5 or $x=-2$x=2 into the expression.

Looking at the second term, we have the rational expression $\frac{5}{x^2+3x+2}$5x2+3x+2, so we can't have the denominator $x^2+3x+2=0$x2+3x+2=0. We can solve this equation as follows.

$x^2+3x+2$x2+3x+2 $=$= $0$0  
$\left(x+1\right)\left(x+2\right)$(x+1)(x+2) $=$= $0$0 Factorise the quadratic
$x$x $=$= $-1$1,$-2$2 Set each factor equal to zero to get the solutions

Hence, we can't substitute $x=-1$x=1 or $x=-2$x=2.

Do: Since $x=-2$x=2 was already forbidden in the first term, we only have three forbidden substitutions, $x=-5$x=5,$-1$1,$-2$2.


Question 2

Write the set of values for which the rational expression $\frac{\left(m-2\right)\left(m+5\right)}{m-10}$(m2)(m+5)m10 is definable.


We can see that the denominator $m-10$m10 cannot be equal to zero, so $m=10$m=10 is not allowable.

Recall that we can use set notation to write the set 'All values of $m$m except $m=10$m=10' as 

$m$m $:$: $m\ne10$m10 }

which means 'The set of values of $m$m such that $m\ne10$m10'.


Further Examples

Question 3

In the following expression, what values are not allowable as substitutions for the variable $r$r? Write your answer as an equation. If there is more than one value, make sure to enter all appropriate values separated by a comma.


Question 4

In the following expression, what values are not allowable as substitutions for the variable $u$u? Write your answer as an equation. Make sure to enter all appropriate values, separated by a comma if there is more than one value.


Question 5

Which of the following is the entire possible set of values that $y$y can take in the rational expression $\frac{2y-3}{\left(4y-8\right)\left(y+7\right)}$2y3(4y8)(y+7)?

  1. { $y$y $:$: $y\ne\frac{3}{2},2,-7$y32,2,7 }


    { $y$y $:$: $y\ne2,-7$y2,7 }


    { $y$y $:$: $y\ne-3,-8,7$y3,8,7 }


    { $y$y $:$: $y\ne8,-7$y8,7 }


    { $y$y $:$: $y\ne\frac{3}{2},2,-7$y32,2,7 }


    { $y$y $:$: $y\ne2,-7$y2,7 }


    { $y$y $:$: $y\ne-3,-8,7$y3,8,7 }


    { $y$y $:$: $y\ne8,-7$y8,7 }




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