I have a systems of equations for a website that relies on the solution. I have just read up on the subject online but I still can't come to a conclusion. The equation is this:
$$8000 \lt xy^5 \lt 8100$$ $$xy^4 - xy^3 \gt 5000$$
I have no idea how to solve for the greater then or less then part of the equation. I also have no idea how to solve for exponents and finally if it even can be solved! all help is greatly appreciated! thank you for your answer!
The system you have is not solvable.
Since we don't know the exact values of $xy^5$ and $xy^4-xy^3$, let's put in some values, say $r$ and $s$, and see what happens. We have: $$xy^5 = r \tag{1}$$ $$xy^4 -xy^3 = s \tag{2}$$ $$8000 < r < 8100$$ $$s > 5000$$
We know that $y \not= 0$, since otherwise $r$ would equal zero, contradicting the fact that $r > 8000 > 0$. Thus, from (1), we can find $$x = \frac{r}{y^5} \tag{3}$$
Substituting (3) into (2) gives $$ry^{-1} - ry^{-2} = s$$ which we can multiply by $y^2$ to get $$ry - r = sy^2$$ $$0 = sy^2 - ry + r \tag{4}$$
I'm not sure if you're familiar with the quadratic equation, but we can use the discriminant to tell when (4) will have solutions. Specifically, (4) will have (real) solutions when (and only when!) $$r^2 - 4rs > 0$$ But, we know that $s > 5000$, so $$r^2 - 4rs < r^2 -4r(5000)$$ $$r^2 - 4rs < r(r-20000)$$ But, for $8000 < r < 8100$, the right hand side is always negative, so our discriminant is always less than zero, and we have no real solutions.