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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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    Try to write this better using LATEx,couldn´t understand completely...in the right side of equality is 26?2012-07-24
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    Substitute $u=x-1$ and $v=y+2$?2012-07-24
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    @Auke: ITYM $u = (x-1)^{-1}$ and $v = (y+2)^{-1}$.2012-07-24
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    @JohannesKloos I already tried it, and so did my math teacher, couldn't figure it out2012-07-24
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    With Johannes' substitution. you'd get $$\eqalign{3u+4v&=2\cr6u-7v&=-3}.$$ Multiply the first equation by $-2$ and add to the second. This will give the equation $-15v=-7$. Then $v=7/15$; whence $y={15\over7}-2=1/7$. Now solve for $u$ and then $x$.2012-07-24
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    Thanks @JohannesKloos, that works better :)2012-07-24
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    @DavidMitra So I did that but I got pretty confused. So when its $$(y+2)^{-1} = \dfrac 7{15}$$ why do you have to change the $\frac 7{15}$ to $\frac {15}7$?2012-07-24
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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

1 Answers 1

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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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    Could be, it was just a quick-n-dirty calculation. Edit: fixed.2012-07-24
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    I used the elimination method for most of these, not linear combination, hope my teacher won't take points off2012-07-24
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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