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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!

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    Yes, A is a in my edit2012-09-03

1 Answers 1

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For the first note ${\frac {1}{a-b}-\frac {1}{a+b}}=\frac {2b}{a^2-b^2}$ so you have $2abx-2aby=2(a^2-b^2)$

For equation 2 (it is really two equations) you should split them apart into $6x+5y=7x+3y+1$, which means $2y=x+1, y=\frac {x+1}2$ and $6x+5y=2x+12y-2, 4x=7y-2$

For Equation 3 (though you have two equations) the basic substitution method works. From the first $x=-\sqrt \frac 32 y$, plug that into the second and you have an equation in $y$.

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    @aayush: You *should* be confused about the first one, as it's one equation, which you're supposed to solve for two variables--impossible to solve using basic substitution only. For the second and third, Ross has pretty much set you up completely. All you've got to do is substitute. For the second, I'd change the first conclusion equation to $x=2y-1$, substitute $x$ into the second conclusion equation. Then back-substitute to get your $x$-value. The third is even more straightforward.2012-09-03