1)
Start with
Assuming , multiply both equations by y:
Multiply the first equation by 3 and subtract the two equations:
Substitute the value for y in one of the equations to deduce the value for x:
2)
As suggested, we take the LCM and the equations become
Multiply both equations by 6 and you have
From here, you can solve the system as in point 1.
3)
Cross multiplication is exactly the method we used in point 1 to solve the equation: you multiply both equations so that they have the same coefficient for one of the variables, and then add/subtract. For example, if you multiply the first equation by 2 and add the the two equations, you'll simplify the y variable and will be able to solve for x.
4)
The sum and multiplication of rational numbers is rational. So, we only need to prove that is irrational, because:
is the sum between 5, that is rational, and
. So, if this sum is irrational, is must be because of
. So, if this product is irrational, it must be because of
Here's the proof that is irrational, by contradiction:
Here's the comment: we start by assuming that we can write as the ratio between a and b, two integers with no common factors. We square both sides, and find you that a^2 is a multiple of 3, so a itself must be a multiple of 3. If we write a as 3k, we find out that b^2 is also a multiple of 3, and so is b.
We started assuming that a and b had no common factors, but we found out that they are both multiple of 3, hence the contradicition.
5)
Every integer can be a multiple of six, or be 1,2,3,4 or 5 units away from a multiple of 6. In fact, if a number is 6 units away from a multiple of six, it is the next multiple of 6, and the count restarts.
This means that we can write every integer in one of these forms:
A multple of 6 is even, so if we sum another even number we'll get an even result:
The remaining cases are the sum of an even number (6q) and an odd one (1, 3 or 5):