Question

First will be marked as brainliest

Answer 1

1)

Start with

$\begin{cases}x+\frac{6}{y} = 6\\3x-\frac{8}{y}=5\end{cases}$

Assuming $y\neq 0$, multiply both equations by y:

$\begin{cases}xy+6 = 6y\\3xy-8=5y\end{cases}$

Multiply the first equation by 3 and subtract the two equations:

$\begin{cases}3xy+18 = 18y\\3xy-8=5y\end{cases}\implies 26=13y \iff y=2$

Substitute the value for y in one of the equations to deduce the value for x:

$x+\frac{6}{2} = 6\iff x+3 = 6 \iff x= 3$

2)

As suggested, we take the LCM and the equations become

$\begin{cases}\frac{3(x+1)+2(y-1)}{6}=8\\\frac{2(x-1)+3(y+1)}{6}=9\end{cases}$

Multiply both equations by 6 and you have

$\begin{cases}3x+3+2y-2}=48\\2x-2+3y+3=54\end{cases} \iff \begin{cases}3x+2y=47\\2x+3y=53\end{cases}$

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 $\sqrt{3}$ is irrational, because:

• $5+2\sqrt{3}$ is the sum between 5, that is rational, and $2\sqrt{3}$. So, if this sum is irrational, is must be because of $2\sqrt{3}$
• $2\sqrt{3}$ is the product of 2, that is rational, and $\sqrt{3}$. So, if this product is irrational, it must be because of $\sqrt{3}$

Here's the proof that $\sqrt{3}$ is irrational, by contradiction:

$\sqrt{3}=\dfrac{a}{b} \iff \dfrac{a^2}{b^2}=3 \iff a^2=3b^2 \iff a=3k\\9k^2=3b^2 \iff b^2=3k^2 \iff b = 3m$

Here's the comment: we start by assuming that we can write $\sqrt{3}$ 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:

$6q,\ \ 6q+1,\ \ 6q+2,\ \ 6q+3,\ \ 6q+4,\ \ 6q+5$

A multple of 6 is even, so if we sum another even number we'll get an even result:

$6q,\ \ 6q+2,\ \ 6q+4\text{ are even}$

The remaining cases are the sum of an even number (6q) and an odd one (1, 3 or 5):

$6q+1,\ \ 6q+3,\ \ 6q+5\text{ are odd}$