Simplify algebraic fractions II

Just as ordinary fractions with particular numbers specified for their numerators and denominators can be written in 'lowest terms' by cancelling common factors, algebraic fractions also can be written in their simplest form by removing common factors in the same way.

The following list of algebraic fractions is a list of 'equivalent fractions'. Whatever numbers are substituted for the letters, the fractions are all the same.

$\frac{a}{3b}$a3b​

$\frac{a^2+a}{3ab+3b}$a2+a3ab+3b​

$\frac{ax+ay}{3bx+3by}$ax+ay3bx+3by​

$\frac{(a+1)^2-1}{3ab+6b}$(a+1)2−13ab+6b​

We generated these equivalent fractions by multiplying the numerator and denominator of the algebraic fraction $\frac{a}{3b}$a3b​ by the same expression.

To reverse the process and arrive at the simplest form of the fraction, we factorise the numerators and the denominators to discover any common factors which can then be cancelled.

Example 1

Simplify the last fraction in the above list by factorising the numerator and the denominator.

$\frac{(a+1)^2-1}{3ab+6b}$(a+1)2−13ab+6b​	$=$=	$\frac{(a+1+1)(a+1-1)}{3b(a+2)}\ \ \ \text{(difference of two squares in the numerator)}$(a+1+1)(a+1−1)3b(a+2)​   (difference of two squares in the numerator)
	$=$=	$\frac{(a+2)a}{3b(a+2)}$(a+2)a3b(a+2)​
	$=$=	$\frac{a}{3b}$a3b​

Strictly speaking, the two fractions $\frac{(a+1)^2-1}{3ab+6b}$(a+1)2−13ab+6b​ and $\frac{a}{3b}$a3b​ are not quite the same. They are the same provided $a\ne-2$a≠−2. If we substitute $a=-2$a=−2 in the first fraction we obtain $\frac{0}{0}$00​ which has no meaning and we would say the fraction is undefined when $a$a has this value. However, we can have $a=-2$a=−2 in the second fraction.

 

Example 2

Simplify $\frac{x^2+5x+4}{x^2-2x-3}$x2+5x+4x2−2x−3​.

We factorise the numerator and the denominator to obtain $\frac{(x+1)(x+4)}{(x+1)(x-3)}$(x+1)(x+4)(x+1)(x−3)​. The factor $x+1$x+1 is common. So, we cancel it to see that

$\frac{x^2+5x+4}{x^2-2x-3}=\frac{x+4}{x-3}$x2+5x+4x2−2x−3​=x+4x−3​

This equivalence is correct except when $x=-1$x=−1 for which case the original fraction was undefined.

 

Example 3

Simplify $\frac{x^2+xy-2y^2}{x^2+ax-xy-ay}$x2+xy−2y2x2+ax−xy−ay​.

The numerator has factors that can be found by guessing and checking, or we might notice that if $x=y$x=y then $x^2+xy-2y^2=0$x2+xy−2y2=0. This means that $x-y$x−y must be a factor. Then, it is not hard to see that the other factor is $x+2y$x+2y.

The denominator can be factorised by grouping the terms:

$x^2+ax-xy-ay$x2+ax−xy−ay	$=$=	$x(x+a)-y(x+a)$x(x+a)−y(x+a)
	$=$=	$(x+a)(x-y)$(x+a)(x−y)

Thus, we have $\frac{x^2+xy-2y^2}{x^2+ax-xy-ay}=\frac{(x-y)(x+2y)}{(x+a)(x-y)}$x2+xy−2y2x2+ax−xy−ay​=(x−y)(x+2y)(x+a)(x−y)​ and so, the simplified fraction is $\frac{x+2y}{x+a}$x+2yx+a​, provided $x\ne y$x≠y.

Worked examples

Question 1

Question 2

Question 3