# Whats the best way of solving this ratio problem?

I have a question relating to a school report that I submitted, essentially I was asked to prove why a certain value ($$x_1$$) had a large increase (40%) as compared with a second value ($$x_2$$). I wanted to prove it using the ratio of a second variable $$y$$.

Lets say the function is, $$x = y + C$$ where $$y$$ is a variable and $$C$$ is a constant.

e.g. lets say there are two cars, and their speed ($$x$$) is described as a function of car engine size ($$y$$) and a drag coefficient ($$C$$), such that: $$x = y + C$$. I want to describe that the ratio of change in $$x$$ of car1 and car2, i.e. $$x_2/x_1$$ is equal to the ratio of the cars engine sizes $$y_2/y_1$$ added to a constant.

How do i express that the ratio of $$x$$ is equal to the ratio of $$y + C$$. I was thinking about writing it as: $$\frac{x_2}{x_1} = \frac{y_2 + C}{y_1 + C}$$

or possibly; $$\Delta x = \frac{y_2 + C}{y_1 + C}$$ Furthermore, I wondered if there's a simple way to relate $$\Delta x$$ and $$\Delta y$$ such that; $$\Delta x = k \Delta y$$ I went and did some research on some of my old math, and realized that: $$\Delta x = x_2 - x_1$$

So whats the best way to express ratios using algebra?

• $\Delta x = x_2-x_1 = (y_2+C)-(y_1+C) = y_2-y_1 = \Delta y$. However, I would call this the change in speed, not ratio of speeds. Ratio of speed would simply be the first expression you wrote. Also, assuming $y_2>y_1$, and $C$ positive, $x_2/x_1$ would always be less than $y_2/y_1$. I dont think anything more can be said about this generally – Anvit Dec 5 '18 at 8:06