How to solve simultaneous equations of the form $\frac{a}{x}+\frac{b}{y}=1$ really fast? Context:

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*Taking the major and minor axes of an ellipse as the $x$ and $y$ axes respectively find the equation of the ellipse passing through the points $(1,\sqrt{6})$ and $(3,0)$.

*Taking the major and minor axes of an ellipse as the $x$ and $y$ axes respectively find the equation of the ellipse passing through the points $(2,4)$ and $(5,\sqrt{2})$.

In problems like these, I frequently find myself solving simultaneous equations of the form $\frac{a}{x}+\frac{b}{y}=1$. For example, when doing math 2, I had to solve the following system of equations (I assumed $p=a^2$, $q=b^2$).
$$\frac{4}{p}+\frac{16}{q}=1\tag{1}$$
$$\frac{25}{p}+\frac{2}{q}=1\tag{2}$$
I'll now show how I solved them.

$$\frac{4}{p}+\frac{16}{q}=1\tag{1}$$
$$\frac{25}{p}+\frac{2}{q}=1\tag{2}$$
Now in $(1)$,
$$\frac{4q+16p}{pq}=1$$
$$4q+16p=pq$$
Again in $(2)$,
$$\frac{25}{p}+\frac{2}{q}$$
$$\frac{25q+2p}{pq}=1$$
$$25q+2p=pq$$
Now,
$$4q+16p=25q+2p$$
$$21q-14p=0$$
$$21q=14p$$
$$q=\frac{14p}{21}$$
Now,
$$\frac{4}{p}+\frac{16}{q}=1$$
$$\frac{4}{p}+\frac{16}{\frac{14p}{21}}=1$$
$$\frac{4}{p}+\frac{16}{\frac{14p}{21}}=1$$
$$\frac{4}{p}+\frac{336}{14p}=1$$
$$\frac{56+336}{14p}=1$$
$$14p=392$$
$$p=28$$
Now,
$$\frac{4}{28}+\frac{16}{q}=1$$
$$\frac{1}{7}+\frac{16}{q}=1$$
$$\frac{16}{q}=\frac{6}{7}$$
$$6q=112$$
$$q=\frac{56}{3}$$
So, the equation of the ellipse is $$\frac{x^2}{28}+\frac{y^2}{\frac{56}{3}}=1\ \text{(Ans.)}$$

As you can see, the process is pretty long and time-taking. Is there any way to find the values of $p,q$ in 2-3 lines?

Use the substitution $x = \dfrac{1}{p} , y = \dfrac{1}{q} $, then your equations become:
$ 4 x + 16 y = 1 $
$ 25 x + 2 y = 1 $
And these can solved very easily, using for example, elimination
$(1) - 8 \times (2) $:
$- 196 x = - 7 \Rightarrow x = \dfrac{7}{196} \Rightarrow p = \dfrac{196}{7} = 28 $
And then, from equation $(2)$,
$y = \dfrac{1}{2} ( 1 -  25 x ) = \dfrac{3}{56} $
Therefore, $q = \dfrac{56}{3}$
