Question

I really need help with multiplying fractions!

Answer 1

proper or improper fractions?

Answer 2

What is the trouble?

Answer 3

a/b * c/d = (a*c)/(b*d)

Answer 4

Can you give some examples? :)

Answer 5

(x-y/x^2-1)(x-1/x^2-y^2)

Answer 6

So what's the problem, then...?

Answer 7

\[\frac{a}{b}*\frac{c}{d}=\frac{a*c}{b*d}\] so an example would be \[\frac{2}{3}*\frac{5}{4}=\frac{10}{12}\]

The same applies for your algebraic example
\[\frac{x-y}{x^2-1}\frac{x-1}{x^2-y^2}\] however you can make life simpler for yourself by looking common factors. For instance in the example I gave above\[\frac{2}{3}*\frac{5}{4}\] there is a common factor of 2 which you can cancel out. that is \[\frac{2}{3}*\frac{5}{4}=\frac{2*5}{3*4}=\frac{1*5}{3*2}=\frac{5}{6}\]. So in your algebraic expression, look to see what you factorise, in other words that \[x^2-1=(x-1)(x+1)\] and \[x^2-y^2=(x-y)(x+y)\] So we now have \[\frac{x-y}{x^2-1}\frac{x-1}{x^2-y^2}= \frac{(x-y)}{(x-1)(x+1)}*\frac{(x-1)}{(x-y)(x+y)}=\frac{(x-y)(x-1)}{(x-1)(x+1)(x-y)(x+y)}\] Now the (x-1) appears on both the numerator and denominator, as does the (x-y) term, so these cancel to give you and answer of \[\frac{1}{(x+1)(x+y)}\]

I hope this helps explain it!

Answer 8

thank you!