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I'm really not good at math.

I have two types of values, we can call them X and Y.

  • When X is 16, Y is 1.5.
  • When X is 24, Y is 1.25.
  • When X is 32, Y is 1.

Now I want to find the Y for when X is 20, for example.

I've tried a few times to put up a formula like 16x = 1.5y but I never get anywhere.

Be aware that when X increases, Y decreases.

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    $\begingroup$ Hint: assume (for the moment) that $y=mx+b$ for suitable constants, $m,b$. You only need two of the given relations to solve for $m,b$...but is the third relation satisfied? If yes, then you are done. If no, then you need a different sort of functional form. $\endgroup$
    – lulu
    Jan 24 '20 at 15:32
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    $\begingroup$ Should note: there are infinitely many functions that satisfy those three relations so there is no sense in which you can determine $y$ as a unique function of $x$. Best you can do is to assume some functional form. $\endgroup$
    – lulu
    Jan 24 '20 at 15:33
  • $\begingroup$ @lulu I added 3 numbers, instead of 2 to show that they are all in a streight line. In theory this line can probably turn at a later point, but in this case, it does not. $\endgroup$ Jan 24 '20 at 15:37
  • $\begingroup$ @JensTörnell We can actually uniquely determine a line from just $2$ points. The problem is when some of your points aren't on the line, in which case you can use different types of curves or find a line that is 'close' to what you want. $\endgroup$
    – Jam
    Jan 24 '20 at 16:11
  • $\begingroup$ It appears that as $X$ increases by $8$, $Y$ decreases by $0.25$. Use that information to determine what $Y$ would be when $X=0$. Do you see why $aX = bY$ would be too simple a model? $\endgroup$
    – John Joy
    Jan 25 '20 at 0:52
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These three points fit a straight line, as you can confirm by plotting them. A spreadsheet will do a regression and give the equation. You can use the two-point form $$Y-1.5=\left(\frac{1.25-1.5}{24-16}\right)\cdot (X-16)$$ and simplify it to $$Y=-\frac X{32}+2$$

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The simplest answer is $Y=2-\frac X{32}$, which will give that, when $X=20$, $Y=1.375$.

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First, you should be able to tell if $X$ and $Y$ are, indeed, linearly related as you intend. Now, assuming this is the case, we use the formula for finding the straight line that passes through point $(x_{1},y_{1})$ and $(x_{0},y_{0})$ which is given by: $$ y-y_{0}=\frac{y_{1}-y_{0}}{x_{1}-x_{0}}(x-x_{0})$$ Taking $x_{0} = 32, y_{0} = 1, x_{1}=16$ and $y_{1}=1,5$ we get: $$y - 1 = -\frac{1}{32}(x-32)$$ Note that the missing point $(24,1.25)$ also belongs to this straight line, since if $x = 24$ then: $$ y - 1 = \frac{8}{32} \Rightarrow y = 1.25$$.

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    $\begingroup$ Good explanation. A note for the question asker: The letters with subscripts ($x_0,x_1,y_1,y_1$) are known numbers like your value of $32$. But the letters without subscripts ($x,y$) are variables, which can take on different values (e.g., when you want to know what $y$ is, when $x$ is a certain value). See more here. $\endgroup$
    – Jam
    Jan 24 '20 at 16:16
  • $\begingroup$ Yes! Thanks for the comment! $\endgroup$
    – IamWill
    Jan 24 '20 at 16:17
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We can fit a linear relationship to your existing data.

Whenever $x$ increase by $8$, $y$ reduces by $0.25$, that is the gradient is $\frac{-0.25}{8}$.

Try to simplify the following: $$\frac{y-1}{x-32}=\frac{-0.25}{8}$$

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