# Calculating the price of product before increase in price by 30%?

Problem to solve:

To buy 2 products the seller gives 30% discount for the less expensive product. The customer/buyer payed for both products 300$. What is the biggest/max price of the less expensive product before the 30% discount? What I did: let x be the first product let y be the second product which is lowered by 30%, the less expensive product First equation:$x+y-(y*0.3)=300{$}$

Second equation ? I tried but just did a modification of the previous one which is incorrect. Also would derivatives apply here? 

You have shown that $$x+\frac{7}{10}y=300$$ and so $$x=300-\frac{7}{10}y$$ and, in order for $y$ to be less expensive, $$y \lt x$$ by substitution, $$y \lt 300-\frac{7}{10}y$$ $$\frac{17}{10}y \lt 300$$ $$y \lt \frac{3000}{17} \approx 176.48$$ and so there is no real maximum value of $y$, since it cannot equal or exceed $\approx \$176.48$but can be any value arbitrarily close to it and less than it. The inequality $$y \lt x$$ was really the "second equation" that you were looking for, but it isn't an equation. • damn, so easy when you showed it. Tnx. Would overload my mind too much If I hadn't asked the question. Jul 3, 2017 at 16:12 • @eugensunic No problem, don't worry about it. If it helped, just$\checkmark$! :D Jul 3, 2017 at 16:15 • I'm waiting for the interval to pass to +15, already +10 ! Jul 3, 2017 at 16:16 This is an interesting question. Here's a more direct approach. The maximum pre-discount price$y^*$of$Y$(the less expensive product) must be equal to$x$, the price of$X$(the more expensive product). If${y^*}'$is the maximum post-discount price of$Y$, then${y^*}':x=7:10$which gives $$y^*=x=\frac {10}{17}\cdot 300=\color{red}{176.47}$$ • Nice approach, but$\frac{3000}{17} = 176.47\$ to the nearest hundredth. Jul 4, 2017 at 8:53