Sum on simultaneous equation

Question

Solving this simultaneous equation is little bit hard.

The equation $$\frac{1}{y-2} - \frac{1}{x+2}=\frac{1}{60}$$ and $$\frac{1}{x} - \frac{1}{y}=\frac{1}{60}.$$ How to solve it please explain.

Answer 1

It's $$xy=62x-62y+244$$ and $$xy=60y-60x,$$ which gives $$62x-62y+244=60y-60x$$ or $$y=x+2$$ and I hope the rest is smooth.

I got $(10,12)$ and $(-12,-10)$.

Answer 2

from the second equation we get by multiplication with $$xy\ne 0$$ $$y-x=\frac{1}{60}xy$$ solving for $x$ we obtaine $$x=\frac{y}{\frac{1}{60}y+1}=\frac{60y}{y+60}$$ plugging this in the first equation we obtain 8after symplifications $$61y^2+122y+120=0$$ solving this we obtain $$x=-12,y=-10$$ or $$x=10,y=12$$

Answer 3

Given equations: \begin{equation} \dfrac{1}{y-2} - \dfrac{1}{x+2} = \dfrac{1}{60} \tag{1} \end{equation} and \begin{equation} \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{1}{60} \tag{2} \end{equation}

From (1), we have \begin{align} & \dfrac{1}{y-2} - \dfrac{1}{x+2} = \dfrac{1}{60} \notag \\ \implies & \dfrac{(x + 2) - (y-2)}{(y-2)(x+2)} = \dfrac{1}{60} \notag \\ \implies & 60 x - 60 y +240 = xy +2y -2x -4 \tag{by cross multiplication} \\ \implies & xy + 62y - 62 x -244 = 0 \tag{3} \end{align} Furthermore, from (2) we have \begin{align} & \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{1}{60} \notag \\ \implies & \dfrac{1}{x} = \dfrac{1}{60} + \dfrac{1}{y} = \dfrac{y + 60}{60 y} \notag \\ \implies & x = \dfrac{60y}{y + 60} \tag{4} \end{align} Putting the value of $x$ from (4) to (3) yields, \begin{align} & \left( \dfrac{60 y}{y + 60}\right) \times y + 62 y - 62 \times \left( \dfrac{60 y}{y + 60}\right) - 244 = 0 \notag \\ \implies &60 y^{2} + 62 y(y + 60) - \left(62 \times 60\right) y - 244 (y + 60) = 0 \notag \\ \implies & 122 y^{2} - 244 y - (244 \times 60) = 0 \notag \\ \implies & y^{2} - 2y - 120 = 0 \notag \\ \implies & y^{2} - 12y + 10y - (12 \times 10) = 0 \notag \\ \implies & y (y - 12) + 10 (y - 12) = 0 \notag \\ \implies & (y + 10) (y - 12) = 0 \notag \\ \implies & y = -10, 12 \tag{5} \end{align} Using (4) and (5), the value of $x$ can be found as, \begin{align} x = \dfrac{60 \times (-10)}{-10 + 60}, \, \dfrac{60 \times 12}{12 + 60} = -12, 10 \end{align}