I've spent hours on this and I keep getting mixed answers. I need to know the rules for multipling, dividing, adding, subtracting equations involving fractions. I google search but the information is not organized, I get sources telling me to find a common denominator, to multiply by the inverse, to just combine the fractions and all these stuff. If someone can please guide me its pretty elementary maths but it would help alot here are some of the questions I should be able to do:
I used to be able to do these but I just forgot since it's been a while. If I can do these, I can be able to do anything. If you are going to show please explain using words the steps you did it would help alot thanks. Relevant khanacademy links would be appreciated (could not find them there).
Much thanks.
First two: $$\frac{a}{b} \times \frac{c}{d}=\frac{a\times c}{b \times d}$$ In the first case, the numerator of the first fraction and the denominator of the second also involves adding/subtracting, so you have to use parentheses: the answer of the first will then be: $$\frac{(4a-12) \times 10b}{5b^3 \times (a^3-27)}$$
You can write this as: $$\frac{(4a-12) \times 10}{5b^2 \times (a^3-27)}\times \frac{b}{b}$$ And of course, $\frac{b}{b}=1$. Note that this may be simplified further, but this is left as an exercise.
The second one: dividing is multiplying with its reciprocal: $$\frac{a}{b} \div \frac{c}{d}=\frac{a}{b} \times \frac{d}{c}$$
The last two are somewhat harder. With adding and subtracting fractions, you need to have the same denominator: $$\frac{a}{b}+\frac{c}{d}=\frac{a}{b} \times \frac{d}{d}+\frac{c}{d} \times \frac{b}{b}=\frac{a\times d}{b \times d}+\frac{c \times b}{d \times b}=\frac{a\times d+c\times b}{b\times d}$$
In the third, the fraction on the left has a more complicated denominator, so we'll try to turn the right denominator into $x^2-1$.
Remember that $(a-b)(a+b)=a^2-b^2$. In this case, $(x+1)(x-1)=x^2-1$, so we've got to multiply the right fraction with $(x-1)$: $$\frac{1}{x^2-1}-\frac{2\times (x-1)}{x^2-1}=\frac{-2x-1}{x^2-1}$$
The second is about the same: we have to multiply something with $(k+3)$ to get $k^2+2k-3$. It's $(k-1)$, because $(k-1)(k+3)=k^2+2k-3$. Then it's just adding the numerators.