If the area of a rectangular field is $2,\!000\,\mathrm{m^2}$ and it’s $10\,\mathrm{m}$ longer than it is wide, how long is the field? Questions like this were of interest to the ancient Babylonians. Show assumed knowledge
We might decide to call the length and width $x$ and $y$ respectively ($l$ and $w$ would also be good choices for the variable names). Then the information we have is: \begin{align*} xy &= 2000\tag{$1$}\\ x &= y + 10\tag{$2$} \end{align*} We can substitute equation 2 into equation 1, obtaining: \begin{align*} (y + 10)y &= 2000\\ y^2 + 10y - 2000 &= 0\\ y &= \frac{-10\pm\sqrt{100 -4\times(-2000)}}{2}\\[5pt] &= \frac{-10\pm\sqrt{8100}}{2}\\[5pt] &= \frac{-10\pm 90}{2}\\[5pt] &= -50,40 \end{align*} A width can’t be negative, so we can discard the first solution. If $y=40$ then $x=40+10=50$ so the field is $50\,\mathrm{m}$ long.
Alternatively, you might have solved equation 2 for $y$ so we can get rid of the $y$ in the first equation and directly find $x$, the length. \begin{align*} y &= x - 10\\ x(x-10) &= 2000\\ x^2 - 10x - 2000 &= 0\\ x &= \frac{10 \pm\sqrt{100 - 4\times(-2000)}}{2}\\[5pt] &= \frac{10\pm\sqrt{8100}}{2}\\[5pt] &= \frac{10\pm 90}{2}\\[5pt] &= -40,50 \end{align*} Again we can discard the negative solution, so the length of the field is $50\,\mathrm{m}$.
It’s also possible that you solved equation 1 for $y$: \[y = \frac{2000}{x}\] and then substituted that into equation 2: \[x = \frac{2000}{x} + 10\] then multiplied both sides by $x$ to get rid of the fraction \begin{align*} x^2 &= 2000 + 10x\\ x^2 - 10x - 2000 &= 0 \end{align*} and we end up with the same quadratic equation we obtained in the second method.