If it is known that $x>0$, $y>0$, and $z>0$. Let $x+y=2z$ and let $4y+z=3x$. If $k,w, $ and $p$ are positive integers, find the smallest possible value of $k$ such that $kx=wy$ and $kx=pz$.
I am not sure where to even begin. Any guidance would be appreciated. I know the answer should be 35, but am unsure as to how to arrive at this answer.
Taking the first two equations, substituting one equation into the other gives $5x=9y$.
Comparing this to $kx=wy$ tells us that there is some integer number $\ell$ such that $k=5\ell$ and $w=9\ell$
Now take $x+y=2z$ to get $5x+5y=10z\Rightarrow 14y=10z$, or $7y=5z$ after dividing by $2$. Comparing this to $wy=pz$ tells us that there is some integer $\ell'$ such that $w=7\ell'$ and $p=5\ell'$. Finally, combining these two results (look at the two expressions for $w$) gives a minimum value with $\ell=7$ and $\ell'=9$.