Little help with some algebra Note: This isn't homework, I'm skipping ahead of class. Please answer all these equations, I'm deathly stuck on them.
Use the substitution method only please. (Find $x$ and $y$.)
\begin{align}
  ax\left({\frac {1}{a-b}-\frac {1}{a+b}}\right)+by\left({\frac {1}{b-a}-\frac {1}{b+a}}\right)=2 
 \end{align}
$a$ is not equal to $b$ or $-b$
Note - It's $ax$ as the first word, I am worried the latex might mess up there.
Equation 2:
\begin{align}
 6x+5y=7x+3y+1=2\left({x+6y-1}\right)
\end{align}
Equation 3:
\begin{align}
 \sqrt{2}x+\sqrt{3}y=0
\end{align}
\begin{align}
 \sqrt{3}x-\sqrt{8}y=0
\end{align}
Thank you for the help!
Note: Please provide the full solution, the more hints i get the more confused i get...

Equation 2:
$$6x+5y=7x+3y+1\Rightarrow 0=x-2y+1 \Rightarrow 2y-1=x$$
$$7x+3y+1=2(x+6y-1)\Rightarrow 5x-9y+3=0$$
Substituting the $\,x\,$ from the first equation you get:
$$5(2y-1)-9y+3=0\Rightarrow y-2=0\Rightarrow y=2\Longrightarrow x=2y-1=3$$  
Equation 3:
$$\sqrt{3}x-\sqrt{8}y=0$$ 
Hence $$\sqrt{3}x=\sqrt{8}y$$ 
So 
$$x=\frac{\sqrt{8}}{\sqrt{3}}y$$ 
Substituting in the second: 
$$\sqrt{2}\frac{\sqrt{8}}{\sqrt{3}}y+\sqrt{3}y=0$$ 
Now you have 
$$\left(\sqrt{\frac{16}{3}}+\sqrt{3}\right)y=0\Longrightarrow y=0\Longrightarrow x=0$$
