Modules for Prerequisite Mathematics

There are no common factors here, so the common denominator is simply the produce of the two denominators. We multiply the numerator and denominator of the first fraction by \(4h\) and the second fraction by \((h+3)\text{.}\) (We need to avoid \(h = 0\) and \(h = -3\) to avoid division by zero).

Then we can put this into one fraction with the common denominator and expand the multiplications in the numerator to see how they can be combined.

Note that the values \(h = 0\) and \(h = -3\) are invalid value for this calculation.

There is a common factor in these denominators, so I'll look for a better choice than just the product of the two denominators. I need \(x^2\) for the first and a factor of \((x-3)\) so the second, so the common denominator \(x^2(x-3)\) should work. I'll multiply the numerator and denominator of the first fraction by \((x-3)\text{,}\) and by \(x\) for the second fraction.

I'll expand the various multiplications and write the result over the common demoninator.

I could write this numerator different, but it's not obviously simpler one way of another. I'll leave it like this. Note that \(x = 0\) and \(x = 3\) are invalid value for this calculation.

There are no common factors here, so the common denominator is simply the produce of the two denominators. We multiply the numerator and denominator of the first fraction by \(t^3\) and the second fraction by \((\tan t)\text{.}\)

Then we can put this into one fraction with the common denominator and expand the multiplications in the numerator to see if they can be combined.

There is nothing else to be done to simplify or combine this expression. Note that the value \(t = 0\) and all the possible (infintely many!) \(t\) such that \(\tan t = 0\) are invalid for this calculation.

Both fractions have powers of \((j - 7)\) in the denominators. I should be able to use the highest power of \((j - 7)\) as the common denominator. In this plan, I actually can leave the second fraction entirely untouched and multiply the numerator and denominator of the first fraction by \((j - 7)\text{.}\)

Then I can expand the numerator in the first fraction and write the subtraction over the common denominator. I must be careful to apply the subtraction to both piece of the second numerator.

Then I can simplify the numerator by grouping like powers of \(j\text{.}\) The \(j^2\) terms will cancel.

There is nothing more to do to simplify this expression. Note that \(j = 7\) is invalid for this calculation.

First we can simply multiply the numerators and denominators.

I notice that there are some common factor here. Since everything here is multiplication, everything is already a factor of the whole numerator or denominator, so I can proceed to cancel some terms. First, \(z^2\) is a factor of both numerator and denominator, so I can cancel that off.

Next, \((z-2)\) is also a common factor, so I can cancel it off as well.

INow there are no common factors, so this is the final result. Throughout the calculation, I note that \(z = 0\) and \(z = 2\) are values that would lead to division by zero and therefore must be avoided.

To divide fraction, we multiply by the reciprocal.

Then we can try to simplify. There is a common factor of \(x^2\) in the numerator and denominator, so we can cancel that off.

Looking more carefully, we can factor \(7\) out of the numerator and \(4\) out of the denominator.

This factoring reveals a common factor of \((x-2)\) in the numerator and denominator. We can cancel this common factor off as well.

No more simplification can be done here, since this is a pretty concise answer. Note that throughout this question, we have to avoid \(x = 2\) and \(x = 0\text{.}\)