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Solve system 1/x - 1/y = 1, 2x + y = 7xy.

Jun 2, 2020

#1
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Solve system  $$\dfrac{1}{x} - \dfrac{1}{y} = 1,\ 2x + y = 7xy \quad| \quad x \ne 0,\ y\ne 0$$.

$$\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{x} - \dfrac{1}{y}} &=& \mathbf{1} \quad| \quad *xy \\\\ \dfrac{xy}{x} - \dfrac{xy}{y} &=& xy \\\\ y - x &=& xy \\ y &=& xy +x \\ y &=& x(y +1) \\\\ \mathbf{x} &=& \mathbf{\dfrac{y}{y+1}} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{2x+y} &=& \mathbf{7xy} \quad | \quad \mathbf{x=\dfrac{y}{y+1}} \\\\ 2*\dfrac{y}{(y+1)}+y &=& 7*\dfrac{y}{(y+1)}*y \\\\ \dfrac{2y}{(y+1)}+y &=& \dfrac{7y^2}{(y+1)}\quad| \quad *\dfrac{(y+1)}{y} \\\\ 2+y+1 &=& 7y \\ 3 &=& 6y \\\\ y &=& \dfrac{3}{6} \\\\ \mathbf{y} &=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{y}{y+1}} \quad | \quad y=\dfrac{1}{2} \\\\ x &=& \frac{1}{2}\above 1pt \frac{1}{2}+1 \\\\ x &=& \frac{1}{2}\above 1pt \frac{3}{2} \\\\ x &=& \dfrac{1}{2}*\dfrac{2}{3} \\\\ \mathbf{x} &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array}$$

Jun 3, 2020