Algebra equation for percentage increase needed to get the current value as a 20% discount I have some products that I want to increase in value such that a 20% discount gives their current value.  It's been ~25 years since college algebra and so I'm a bit rusty on setting up the equation.
I've been trying to figure out how to solve for X being the percentage increase needed in order that 20% off would give the current value.
For example a product worth 100.  I know a 20% increase would make it 120, but 20% off of that would be 96 which isn't 100.
I'd give a bounty for explaining the algebra and steps to figure it out, but I'm new to this exchange and am unable to award one - thanks if you spend the time to explain this to me!
And if someone wouldn't mind tagging this appropriately for this exchange I'd appreciate it.
 A: Increasing an amount $A$ by $X$ percent means adding $A\cdot \frac{X}{100}$ to $A$, resulting in $A+A\cdot\frac{X}{100}=A\left(1+\frac{X}{100}\right)$.  Decreasing an amount $B$ by $Y$ percent means subtracting $B\cdot\frac{Y}{100}$, resulting in $B-B\cdot\frac{Y}{100}=B\left(1-\frac{Y}{100}\right)$.  To have an $X$ percent increase followed by a $20$ percent decrease with an initial amount $A$, you will first multiply by $1+\frac{X}{100}$ to obtain a new amount.  If we call that amount $B$, then the next step is to decrease $B$ by $20$ percent by multiplying by $1-\frac{20}{100}$.  At this point you will have $A\cdot \left(1+\frac{X}{100}\right)\cdot\left(1-\frac{20}{100}\right)$.  For this to leave you where you started, you need to solve the equation $$A\cdot \left(1+\frac{X}{100}\right)\cdot\left(1-\frac{20}{100}\right) =A.$$  You can cancel $A$ from both sides, leaving an equation $$\frac{4}{5}\left(1+\frac{X}{100}\right)=1$$ with $X$ as the only unknown, which can then be solved by division, subtraction, and multiplication.  Does that get you where you want to be?
