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I have to complete a summer packet of 90 Algebra 2 questions. I have completed 89 of them, the only one I could not get was this. I know the answer is $y = \frac {47}2$, $\frac 17$ according to WolframAlpha, but I have no idea how to reach that answer, my Algebra 2 Honors teacher couldn't figure it out.

Directions: Use substitution or linear combination to solve each system.

$\dfrac 3{(x-1)} + \dfrac 4{(y+2)} = 2$ $\dfrac 6{(x-1)} - \dfrac 7{(y+2)} = -3$

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    @user1036238 We have $v=7/15$. Using the first equation in my previous comment, we have $3u+28/15=2$, or $u={1\over3}\cdot(2-28/15)=2/45$. So $u={1\over x-1}={2\over 45}$. Take reciprocals to get $x-1={45\over2}$. Finally $x={45\over 2}+1={47\over2}$.2012-07-24

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How about this approach: It's not hard to see that $2 \cdot \left(\frac{3}{x-1}\right) = \frac{6}{x-1}$. Therefore, we can do a linear combination approach:

Add $-2$ times the first equation to the second equation. We get:

$ \left(\frac{6}{x-1} - \frac{7}{y+2} \right) - 2 \left( \frac{3}{x-1} + \frac{4}{y+2} \right) = -3 - 2\cdot 2.$

Simplifying gives $\frac{-15}{y+2} =\frac{-7-8}{y+2} + \frac{6-6}{x-1} = -7.$

Can you take it from here?

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    From a mathematical point of view, one method is as good as the other. In practice, linear combination is superior - the method generalizes to more complicated settings (many equations in many variables, coefficients that are something different than rational numbers, ...) and it gives a nice method for computer programming.2012-07-24