Which fraction is more Can anyone help me prove why 
$$1-\left(\frac{x-10}{y}\right) \gt 1-\left(\frac{x}{y+10}\right)$$ 
when $y < x$.
I have no idea how to show this algebraically so i'd really like some guidance.
 A: Combine the fractions with the leading $1-$
$$1-\left(\frac{x-10}{y}\right) \stackrel{?}{\gt} 1-\left(\frac{x}{y+10}\right)
\\\left(\frac{y-x+10}{y}\right) \stackrel{?}{\gt} \left(\frac{y+10-x}{y+10}\right)$$
Now the numerators are the same.  We can see that there is no clear comparison between them.  
If $y-x+10 \gt 0$ and $y \gt 0$ the left will be larger because both are positive and the left denominator is smaller.  
If $y-x+10 \lt 0$ and $y \gt 0$ the left will be smaller because both are negative and the left denominator is smaller.  
And other cases depending on $y, y+10$.
A: We'll try $x=-11$ and $y=-7$. Now
$$-2=1−\frac{x−10}{y}> 1-\frac{x}{y+10}=\frac{14}{3}$$
So actually, your statement is false.
A: Simplifying the expressions, we get the equivalent inequality
$$
\frac{y-x+10}y\gt\frac{y+10-x}{y+10}
$$
This inequality is true if $y\not\in\{0,10,x-10\}$ and one and only one of the following are true
$$
\begin{array}{rcl}
\small{\text{1.}}&x\lt y+10&\text{both numerators are positive}\\
\text{2.}&-10\lt y\lt0&\text{denominators have different signs}
\end{array}
$$
Unfortunately, there are many values of $(x,y)$ in which neither is true, e.g. $(12,1)$, and many in which both are true, e.g. $(1,-1)$. Therefore, without some other restriction on $(x,y)$, the inequality is not true in general.
