1000 - 10% = 900
900 + 10% = 990
Is there a way when the first example always result in the second value (990), independent of the value? I can change the percentage with no problem.
What I really need is the result of the second calc to be 1000 (or whatever other value of input).
On
Hint:
You want: $$ \left[1000(1-x)\right](1+x)=1000 $$
solve the equation and you can see that it is possible only for $x=0$
On
You want to solve the equation $x-px=x+px$.
Observe that $x-px=x+px\iff2px=0\iff[x=0]\vee[p=0]$.
In other words, either the initial amount is zero or the percentage is zero (or both).
On
First of all you should realize that percentage is just a factor, a ratio. Actually $x\% = \frac{x}{100}$ and nothing else. But when you say "$y$ is $p\%$ of $x$" this means $y = \frac{p}{100} x$.
So the expression $1000 - 10\%$ has no exact meaning. If you mean 10% of 1000 it should be written as $$1000 - 10\% \cdot 1000 = 1000 \cdot 100\% - 10\% \cdot 1000 = 1000 \cdot (100\% - 10\%) =1000 * 90 / 100 = 990.$$
Now it's clear you are to solve an equation $1000(1-p)(1+p) = 1000$ but it has no solution appropriate for you, as far as I can judge.
First, you need to write correct equations. When you take $10 \%$ off $1000$ you should write it as $1000(1-10\%)=900$, which can be translated to $1000(1-0.1)=900$ Your first is claiming $1000-0.1=900$ which is not correct. Then if you want to find the percentage to increase $900$ by to get back to $1000$ (I am guessing this is what you want. The question is very difficult to understand), we can do it for a general percentage reduction. If you reduce $1000$ by $x \%$ you get $1000(1-\frac x{100})=y$ If you increase $y$ by $z \%$, you get $y(1+\frac z{100})$. You want this to equal $1000$, which will happen when $1000(1-\frac x{100})(1+\frac z{100})=1000$ or $(1-\frac x{100})(1+\frac z{100})=1$ You can then solve this to give $\frac z{100}-\frac x{100}-\frac {xz}{100}=0$, $$z=\frac x{1-x}$$ When $x$ is small, $z$ will be close to $x$.