We've already had a look at how to multiply and divide algebraic fractions earlier on, but now we can use factorisation to help us solve problems with complex fractions.
Before we go on, let's just review quickly how to multiply and divide algebraic fractions. Doing so is exactly the same as with regular fractions that only involve numbers- we also simplify the problem by cancelling out common factors. To avoid getting confused algebraically, look at common factors of each variable separately.
For example, let's work out what $\frac{6xy}{15y^2}\times\frac{20y}{12x^3}$6xy15y2×20y12x3 is.
I'll start by simplifying the first fraction $\frac{6xy}{15y^2}$6xy15y2, and comparing the numerator and denominator.
I can see that the HCFs of the coefficients and $y$y's are $3$3 and $y$y, respectively. (There is no common factor for the $x$x term). Thus it'll be simplified to $\frac{2x}{5y}$2x5y.
Similarly the second fraction $\frac{20y}{12x^3}$20y12x3 can be simplified to $\frac{5y}{3x^3}$5y3x3.
So now our problem is $\frac{2x}{5y}\times\frac{5y}{3x^3}$2x5y×5y3x3 and we can see that HCFs can also be cancelled out DIAGONALLY across.
$2x$2x and $3x^3$3x3 have an HCF of $x$x, while $5y$5y and $5y$5y can be completely cancelled out.
So our resulting problem is $\frac{2}{1}\times\frac{1}{3x^2}=\frac{2}{3x^2}$21×13x2=23x2.
Sometimes the fractions involved in these multiplication and division problems are too complicated to see their factors straight away, and that's where factorisation comes in. We can use the various factorisation techniques we have learnt previously.
Factorise and simplify the following: $\frac{9x^2}{3xy-6x}\div\frac{3y+9}{y^2+y-6}$9x23xy−6x÷3y+9y2+y−6
Think about how division is just multiplication with the second fraction inverted
Do: So our problem can be rewritten as:
$\frac{9x^2}{3xy-6x}\times\frac{y^2+y-6}{3y+9}$9x23xy−6x×y2+y−63y+9
The denominator of the first fraction can be factorised using HCFs:
$\frac{9x^2}{3xy-6x}$9x23xy−6x | $=$= | $\frac{9x^2}{3x\left(y-2\right)}$9x23x(y−2) | |
$=$= | $\frac{3x}{y-2}$3xy−2 | by cancelling out $3x$3x from top and bottom |
The second fraction can be factorised using the cross method on top and HCFs on the bottom:
$\frac{y^2+y-6}{3y+9}$y2+y−63y+9 | $=$= | $\frac{\left(y-2\right)\left(y+3\right)}{3\left(y+3\right)}$(y−2)(y+3)3(y+3) | |
$=$= | $\frac{y-2}{3}$y−23 | by cancelling out $y+3$y+3 on top and bottom |
So our problem is now:
$\frac{3x}{y-2}\times\frac{y-2}{3}$3xy−2×y−23 | $=$= | $\frac{3x}{1}\times\frac{1}{3}$3x1×13 | by diagonally cancelling out $y-2$y−2 |
$=$= | $\frac{x}{1}\times\frac{1}{1}$x1×11 | by diagonally cancelling out $3$3 | |
$=$= | $x$x |
Factorise and simplify
$\frac{5q}{50pq^2-8p}\times\frac{4pq+24p^2}{q^2+12pq+36p^2}$5q50pq2−8p×4pq+24p2q2+12pq+36p2
Think about how some quadratics don't need to be factorised using the cross method
Do
$\frac{5q}{50pq^2-8p}\times\frac{4pq+24p^2}{q^2+12pq+36p^2}$5q50pq2−8p×4pq+24p2q2+12pq+36p2 | $=$= | $\frac{5q}{2p\left(25q^2-4\right)}\times\frac{4p\left(q+6p\right)}{q^2+12pq+36p^2}$5q2p(25q2−4)×4p(q+6p)q2+12pq+36p2 | using HCF factorisation | ||
$=$= | $\frac{5q}{2p\left(5q+2\right)\left(5q-2\right)}\times\frac{4p\left(q+6p\right)}{q^2+12pq+36p^2}$5q2p(5q+2)(5q−2)×4p(q+6p)q2+12pq+36p2 | using difference of $2$2 squares | |||
$=$= | $\frac{5q}{2p\left(5q+2\right)\left(5q-2\right)}\times\frac{4p\left(q+6p\right)}{\left(q+6p\right)^2}$5q2p(5q+2)(5q−2)×4p(q+6p)(q+6p)2 | ||||
the denominator is a perfect square as $q^2$q2 and $36p^2$36p2 are both squares and $12pq=2\times q\times6p$12pq=2×q×6p | |||||
$=$= | $\frac{5q}{2p\left(5q+2\right)\left(5q-2\right)}\times\frac{4p}{q+6p}$5q2p(5q+2)(5q−2)×4pq+6p | ||||
$=$= | $\frac{5q}{\left(5q+2\right)\left(5q-2\right)}\times\frac{2}{q+6p}$5q(5q+2)(5q−2)×2q+6p | ||||
$=$= | $\frac{10q}{\left(5q+2\right)\left(5q-2\right)\left(q+6p\right)}$10q(5q+2)(5q−2)(q+6p) |
Simplify the following: $\frac{5x+8}{8xy^2}\times\frac{9xy}{25x+40}$5x+88xy2×9xy25x+40
Simplify the following expression:
$\frac{p+7}{5}\times\frac{5p-2}{p^2+14p+49}$p+75×5p−2p2+14p+49
Simplify the following expression:
$\frac{a^2-16}{a\left(a+4\right)}\times\frac{7a+28}{28\left(a-4\right)}$a2−16a(a+4)×7a+2828(a−4)