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...
 A: First Equation:
$$ax\left({\frac {1}{a-b}-\frac {1}{a+b}}\right)+by\left({\frac {1}{b-a}-\frac {1}{b+a}}\right)=2 $$
$$ax\left({\frac {a+b-a+b}{a^2-b^2}}\right)+by\left({\frac {b+a-b+a}{b^2-a^2}}\right)=2$$
$$2abx-2aby=2(a^2-b^2)$$
$$x-y=\frac{a}{b}-\frac{b}{a}$$
From this the obvious solution is $x=a/b \text{ and } y=b/a$. But there will be infinite solutions $x=y+(a/b)-(b/a)$. Just put values for $y$ and get corresponding values for $x$.
Second Equation:
$$6x+5y=7x+3y+1=2(x+6y-1)$$
Break it into $2$ equations
$$6x+5y=7x+3y+1 \text{ and } 6x+5y=2(x+6y-1)$$
$$-x+2y=1 \text{ and } 4x-7y=-2$$
The value of $x$ from the first equation is $x=2y-1$. Put it in second equation.
$$4(2y-1)-7y=-2$$
$$y=2$$
Hence $x=2y-1=3$.
Third equation:
$$\sqrt2x+\sqrt3y=0 $$
From this $x=-\sqrt{3/2}y$. Put this in the second equation.
$$\sqrt3x-\sqrt8y=0$$
$$y(3/\sqrt2 - \sqrt8)=0$$
$$y=0$$
Hence $x=0$.
A: $$ax\left(\frac{1}{a-b}-\frac{1}{a+b}\right)+by\left(\frac{1}{b-a}-\frac{1}{b+a}\right)=2\Longrightarrow$$
$$\frac{ax}{a^2-b^2}(2b)+\frac{by}{b^2-a^2}(2a)=2\Longrightarrow \frac{abx-aby}{a^2-b^2}=1\Longrightarrow$$
$$x-y=\frac{a^2-b^2}{ab}\,\,,\,\,\text{assuming}\,\,ab\neq 0$$
and nothing more can be done as you're only giving one equation with two variables. Of course, the substitution method here cannot possibly work, no matter what your book requires (check this, it may well be you made a mistake while copying the question)
A: 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 
$$\left(\sqrt{\frac{16}{3}}+\sqrt{3}\right)y=0\Longrightarrow y=0\Longrightarrow x=0$$
