I came across a STEP III($2010, Q5$) problem which involves finding the point of intersection between the curve with the equation $$n(m-1)x+m(n-1)y=amn \tag{1}$$ and $$mx+ny=amn \tag{2}$$
where $m$ and $n$ are real constants that satisfy $$0<m<n<1 \tag{3}$$
Equating $(1)$ with $(2)$ yields $$(mn-n-m)(x+y)=0 \tag{4}$$
I then simply solved for $x$ and $y$ simultaneously in terms of $m$ and $n$.
However, the mark scheme seems to offer a less arduous alternative and is stated as follows:
As the curve $(1)$ and $(2)$ intersect, $$\frac{n}{m} \neq \frac{m(n-1)}{n(m-1)} \tag{5}$$ which yields $$(m-n)(mn-m-n) \neq0 \tag{6}$$ and hence $$(mn-m-n) \neq 0 \tag{7}$$ implying $$x+y=0 \tag{8}$$
But I do not see how the $Eq.(5)$ can be obtained from $(4)$?
The slope of the curves are $\frac{m}{n}$ and $\frac{n(m-1)}{m(n-1)}$.
As these intersect, they aren't parallel, so their slopes are unequal.
So $$\frac{m}{n}\ne\frac{n(m-1)}{m(n-1)} \implies m^2(n-1) \ne n^2(m-1) \implies m^2n-n^2m-m^2+n^2\ne0 \implies (m-n)(mn-m-n)\ne0 \implies(mn-m-n)\ne0 $$ as $m<n \implies m\ne n$