Real Analysis Archimedian Property If $x, y$ are real numbers, with $x &lt; y$, show that $$x &lt; ty + (1-t)x &lt; y$$ for all $t$ such that $ 0 &lt; t &lt; 1$.

I know we need to use Archimedian principle, and I think it may be used twice, but I'm stuck on how to get the extra $x$ in there from $t(y-x) &gt; 1$. 

Any help would be great!
Since $x&lt;y$ and $0&lt;t&lt;1$ so
$$x=tx+(1-t)x&lt;ty+(1-t)x&lt;ty+(1-t)y=y$$
Now let $x&lt;z&lt;y$, define the function $f:[0,1]\rightarrow [x,y]$ with 
$$f(t)=ty+(1-t)x=t(x-y)+x$$
then $f(0)=x$ and $f(1)=y$ and also continuous injective, now mean value theorem implies that there is $t_0$ such that $f(t_0)=z$.
Since $x&lt;y$ and $0&lt;t&lt;1$ so
$$x=tx+(1-t)x&lt;ty+(1-t)x&lt;ty+(1-t)y=y$$
Now let $x&lt;z&lt;y$, define the function $f:[0,1]\rightarrow [x,y]$ with 
$$f(t)=ty+(1-t)x=t(x-y)+x$$
then $f(0)=x$ and $f(1)=y$ and also continuous injective, now mean value theorem implies that there is $t_0$ such that $f(t_0)=z$.