When it comes to working with algebraic fractions and applying the four operations, the process is exactly the same as when we worked with numeric fractions.
Let's have a look at a simple example of multiplying two algebraic fractions.
Simplify $\frac{y}{5}\times\frac{3}{m}$y5×3m
$\frac{y}{5}\times\frac{3}{m}$y5×3m | $=$= | $\frac{y\times3}{5\times m}$y×35×m Multiplying numerator and denominators |
$=$= | $\frac{3y}{5m}$3y5m Simplifying the numerator |
Again, since the numerator $3y$3y and the denominator $5m$5m don't have any common factors, $\frac{3y}{5m}$3y5m is the simplest form of our answer.
Again, the process for dividing is the same as when we divided numeric fractions. We need to multiply by the reciprocal of the second fraction.
Simplify $\frac{m}{3}\div\frac{5}{x}$m3÷5x
$\frac{m}{3}\div\frac{5}{x}$m3÷5x | $=$= | $\frac{m}{3}\times\frac{x}{5}$m3×x5 | Dividing by a fraction is the same as multiplying by its reciprocal. |
$=$= | $\frac{m\times x}{3\times5}$m×x3×5 | Multiply numerators and denominators respectively. | |
$=$= | $\frac{mx}{15}$mx15 |
Again, since the numerator $mx$mx and the denominator $15$15 don't have any common factors, $\frac{mx}{15}$mx15 is the simplest form of our answer.
Simplify the expression:
$\frac{m}{8}\div\frac{3}{n}$m8÷3n
Simplify the following: $\frac{-2x}{11}\div\frac{7y}{5}$−2x11÷7y5
Simplify $\frac{-2x}{11}\div\frac{2x}{3}$−2x11÷2x3.
Sometimes the fractions involved in these multiplication and division problems are too complicated to see their factors immediately, and that's where factoring comes in. We can use the various factoring techniques we have learned previously.
Factor 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 factored using GCFs:
$\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 canceling out $3x$3x from top and bottom |
The second fraction can be factored using the cross method on top and GCFs 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 canceling 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 canceling out $y-2$y−2 |
$=$= | $\frac{x}{1}\times\frac{1}{1}$x1×11 | by diagonally canceling out $3$3 | |
$=$= | $x$x |
Factor 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 factored 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 | ||
$=$= | $\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 | |||
$=$= | $\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}\cdot\frac{9xy}{25x+40}$5x+88xy2·9xy25x+40
Simplify the following expression:
$\frac{p+7}{5}\cdot\frac{5p-2}{p^2+14p+49}$p+75·5p−2p2+14p+49
Simplify the following expression:
$\frac{a^2-16}{a\left(a+4\right)}\cdot\frac{7a+28}{28\left(a-4\right)}$a2−16a(a+4)·7a+2828(a−4)
We want to look at fractions in which either the numerator, denominator, or both contain fractions themselves. There are two techniques for simplifying these more complex fractions. The first is to simplify by rewriting as division, and then multiplying by the reciprocal of the second fraction.
Simplify the expression $\frac{\frac{1}{6}}{\frac{-9}{8}}$16−98.
Firstly, when we have one fraction divided by another, we can multiply the first by the reciprocal of the second.
$=\frac{1}{6}\times\frac{8}{-9}$=16×8−9
Now we want to simplify. We look vertically and diagonally for numbers that have greatest common factors. Doing this we see that $6$6 and $8$8 can be divided by $2$2.
$=\frac{1}{3}\times\frac{4}{-9}$=13×4−9
Now that we've simplified everything we can see, we multiply across horizontally.
$=\frac{4}{-27}$=4−27
Simplify the expression $\frac{-\frac{3}{8}}{-\frac{9}{4}}$−38−94.
Fill in the empty boxes to simplify the expression.
$\frac{\frac{4}{n}}{\frac{5}{n^2}}$4n5n2$=$=$\frac{\editable{}\left(\frac{4}{n}\right)}{\editable{}\left(\frac{5}{n^2}\right)}=\frac{\editable{}}{\editable{}}$(4n)(5n2)=
Simplify the expression $\frac{\frac{28uv^2}{25}}{\frac{49u^2v}{15}}$28uv22549u2v15.