1)Start with[tex]\begin{cases}x+\frac{6}{y} = 6\\3x-\frac{8}{y}=5\end{cases}[/tex]Assuming [tex]y\neq 0[/tex], multiply both equations by y:[tex]\begin{cases}xy+6 = 6y\\3xy-8=5y\end{cases}[/tex]Multiply the first equation by 3 and subtract the two equations:[tex]\begin{cases}3xy+18 = 18y\\3xy-8=5y\end{cases}\implies 26=13y \iff y=2[/tex]Substitute the value for y in one of the equations to deduce the value for x:[tex]x+\frac{6}{2} = 6\iff x+3 = 6 \iff x= 3[/tex]2)As suggested, we take the LCM and the equations become[tex]\begin{cases}\frac{3(x+1)+2(y-1)}{6}=8\\\frac{2(x-1)+3(y+1)}{6}=9\end{cases}[/tex]Multiply both equations by 6 and you have[tex]\begin{cases}3x+3+2y-2}=48\\2x-2+3y+3=54\end{cases} \iff \begin{cases}3x+2y=47\\2x+3y=53\end{cases}[/tex]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 [tex]\sqrt{3}[/tex] is irrational, because:[tex]5+2\sqrt{3}[/tex] is the sum between 5, that is rational, and [tex]2\sqrt{3}[/tex]. So, if this sum is irrational, is must be because of [tex]2\sqrt{3}[/tex][tex]2\sqrt{3}[/tex] is the product of 2, that is rational, and [tex]\sqrt{3}[/tex]. So, if this product is irrational, it must be because of [tex]\sqrt{3}[/tex]Here's the proof that [tex]\sqrt{3}[/tex] is irrational, by contradiction:[tex]\sqrt{3}=\dfrac{a}{b} \iff \dfrac{a^2}{b^2}=3 \iff a^2=3b^2 \iff a=3k\\9k^2=3b^2 \iff b^2=3k^2 \iff b = 3m[/tex]Here's the comment: we start by assuming that we can write [tex]\sqrt{3}[/tex] 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:[tex]6q,\ \ 6q+1,\ \ 6q+2,\ \ 6q+3,\ \ 6q+4,\ \ 6q+5[/tex]A multple of 6 is even, so if we sum another even number we'll get an even result:[tex]6q,\ \ 6q+2,\ \ 6q+4\text{ are even}[/tex]The remaining cases are the sum of an even number (6q) and an odd one (1, 3 or 5):[tex]6q+1,\ \ 6q+3,\ \ 6q+5\text{ are odd}[/tex]