Find the number of pairs of integers (x,y) with x is less than 0, y is less than 10, that satisfy \(\frac{1}{1-\frac{10}{x}} > 1 - \frac{5}{y}.\)
I'm very sorry I haven't learned this...
I asked some people do but they are eating dinner of AFK
oh, no problem! ill just wait and ask other people. thanks you your help though!
Hi Littlemixfan,
Find the number of pairs of integers (x,y) with x is less than 0, y is less than 10, that satisfy
\(\frac{1}{1-\frac{10}{x}} > 1 - \frac{5}{y}\)
First I want to look at formula restrictions.
\(y\ne0\\ x\ne0\\ x\ne10\)
We also have the question restriction of
\(x<0\qquad and \qquad y<10\)
So where can it be so far:
NOW
\(\frac{1}{1-\frac{10}{x}} > 1 - \frac{5}{y}\\ \frac{1}{\frac{x-10}{x}} > 1 - \frac{5}{y}\\ \frac{x}{x-10} > \frac{x-10}{x-10} - \frac{5}{y}\\ \frac{x-(x-10)}{x-10} > - \frac{5}{y}\\ \frac{10}{x-10} > - \frac{5}{y}\\ \frac{x-10} {10}< - \frac{y}{5}\\ - \frac{y}{5}>\frac{x-10} {10} \\ \frac{y}{5}<\frac{-x+10} {10} \\ \frac{y}{1}<\frac{-x+10} {2} \\ y<-0.5x+5 \)
So that is underneath the line y=-0.5x+5
NOTE: Maybe I made an error but I do not think so. Maybe it was meant to be a < sign.
Also,:
I left the > sign in for all my working but it would have been easier just to solve for = and then test the regions afterwards.
So far I have these 2 regions. But since there are 2 regions one where y<0 and one where y>0 I will test both.
Test (-1,+1) and (-1,-1)
\(\frac{1}{1-\frac{10}{x}} > 1 - \frac{5}{y}\\test\;(-1,1)\\ \frac{1}{11}>-4\quad true \\ test\;(-1,-1)\\ \frac{1}{11}>6\quad false \\ \)
So here is the region:
I think there is an infinite number of integer pair solutions.
LaTex:
\frac{1}{1-\frac{10}{x}} > 1 - \frac{5}{y}\\
\frac{1}{\frac{x-10}{x}} > 1 - \frac{5}{y}\\
\frac{x}{x-10} > \frac{x-10}{x-10} - \frac{5}{y}\\
\frac{x-(x-10)}{x-10} > - \frac{5}{y}\\
\frac{10}{x-10} > - \frac{5}{y}\\
\frac{x-10} {10}< - \frac{y}{5}\\
- \frac{y}{5}>\frac{x-10} {10} \\
\frac{y}{5}<\frac{-x+10} {10} \\
\frac{y}{1}<\frac{-x+10} {2} \\
y<-0.5x+5