# How to solve this System of Polynomial Equations?

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? – HipsterMathematician Jul 24 '12 at 20:30
Substitute $u=x-1$ and $v=y+2$? – akkkk Jul 24 '12 at 20:36
@Auke: ITYM $u = (x-1)^{-1}$ and $v = (y+2)^{-1}$. – Johannes Kloos Jul 24 '12 at 20:38
@JohannesKloos I already tried it, and so did my math teacher, couldn't figure it out – user1036238 Jul 24 '12 at 20:39
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 $x$. – David Mitra Jul 24 '12 at 20:43

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. – Johannes Kloos Jul 24 '12 at 20:56
I used the elimination method for most of these, not linear combination, hope my teacher won't take points off – user1036238 Jul 24 '12 at 21:06
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. – Johannes Kloos Jul 24 '12 at 21:29