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Given the system of equations $$ \frac{2}{x} + \frac{3}{y} = 6 \quad \text{and}\quad 5x - y = 4 $$ solve for $x$ and $y$.

I have tried rearranging the equation to substitute either $x$ or $y$, but I wasn't able to solve it. Any help would be appreciated.

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Put $y=5x-4$ in the first equation and multiply the equation by $x(5x-4)$ You will get quadratic equation in $x$. Do you know how to solve a quadratic equation?

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If you substitute $y$ by $5x-4$ in the first equation, you get$$\frac2x+\frac3{5x-4}=6.\tag1$$But\begin{align}(1)&\iff\frac2x+\frac3{5x-4}-6=0\\&\iff\frac{-30 x^2+31 x-8}{x (5 x-4)}=0.\end{align}Can you take it from here?

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  • $\begingroup$ just a question, how did you go from (1) to (2) where you have everything over one denominator? $\endgroup$ – Vecter Feb 13 at 1:55
  • $\begingroup$ I had several fractions and then I summed them into a single fraction. $\endgroup$ – José Carlos Santos Feb 13 at 7:23

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