Suppose that I have the following system of $K+1$ linear equations with $K\geq 3$ $$ \begin{cases} \lambda_j\times \lambda_h'=\lambda'_j\times \lambda_h & \text{ for $K$ different pairs $(j,h)$ taken from $A$}\\ \lambda_1+...+\lambda_K=1 \end{cases} $$
where
the unknowns are $(\lambda_1,...,\lambda_K)$
$(\lambda'_1,...,\lambda_K')$ are known parameters such that $$\lambda'_1+...+\lambda'_K=1$$
$\lambda_1>0,...,\lambda_K>0$ and $\lambda'_1>0,...,\lambda'_K>0$
$A\equiv \{(1,2),(1,3),...,(1,K),(2,K),(3,K),...,(K-1,K)\}$ with cardinality $2K-3$.
Question: could you help me to show that this system has a unique solution that is $$\lambda_1=\lambda'_1,...,\lambda_K=\lambda'_K$$
For example, when $K=5$, we have that $A\equiv \{(1,2),(1,3),(1,4),(1,5), (2,5),(3,5),(4,5)\}$ and a specification of the system could be $$ \begin{cases} \lambda_1\times \lambda_5'=\lambda'_1\times \lambda_5\\ \lambda_2\times \lambda_5'=\lambda'_2\times \lambda_5\\ \lambda_1\times \lambda_4'=\lambda'_1\times \lambda_4\\ \lambda_1\times \lambda_3'=\lambda'_1\times \lambda_3\\ \lambda_3\times \lambda_5'=\lambda'_3\times \lambda_5\\ \lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5=1 \end{cases} $$
which implies $$ \begin{cases} \lambda_2=\lambda_1\times \frac{\lambda_2'}{\lambda_1'}\\ \lambda_3=\lambda_1\times \frac{\lambda_3'}{\lambda_1'}\\ \lambda_4=\lambda_1\times \frac{\lambda_4'}{\lambda_1'}\\ \lambda_5=\lambda_1\times \frac{\lambda_5'}{\lambda_1'}\\ \end{cases} $$ Then, replacing this in $\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5=1 $ we get $$ \lambda_1\times \Big(\overbrace{1+\frac{\lambda_2'}{\lambda_1'}+ \frac{\lambda_3'}{\lambda_1'}+ \frac{\lambda_4'}{\lambda_1'}+ \frac{\lambda_5'}{\lambda_1'}}^{=1/\lambda_1'}\Big)=1 $$ which implies $\lambda_1=\lambda_1'$ and hence $\lambda_2=\lambda_2',\lambda_3=\lambda_3',\lambda_4=\lambda_4',\lambda_5=\lambda_5'$.
I am unable to generalise this procedure to any $K$ elements from the set $A$. Could you help?