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There are two functions: $\frac{32-x}4=y$ and $\frac{48-x}2=y$

For instance, if the two $x$s are $18$ and $24$, the two $y$s will be $3.5$ and $12$. I would like to add the two $x$s and two $y$s: $18+24=42$, $3.5+12=15.5$.

How can I merge the two functions into only one so that I set the $x=42$, I can directly get the $y=15.5$?

lioness99a
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BrotherYao
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  • It appears that you have two completely arbitrary functions and want to "merge" them to get a completely arbitrary result; I can't imagine that there is a reasonable way to do that! – user247327 Feb 24 '20 at 12:21
  • You can't. The pair $18$ and $24$ gives a different answer than for example the pair $24$ and $18$, but you want them to give the same answer. – Jaap Scherphuis Feb 24 '20 at 12:21

3 Answers3

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You have the two functions $$y_1 = (32-x_1)/4 \\ y_2 = (48-x_2)/2$$

Adding them together gives

$$y_1+y_2 = (32-x_1)/4 + (48-x_2)/2$$

which can be simplified to

$$y_1+y_2 = (128-(x_1+2x_2))/4$$

or

$$Y = (128-X)/4$$

where $Y=y_1+y_2$ and $X=x_1+2x_2$.

This is not a function of $x_1+x_2$, which you wanted, but it is a function of $x_1+2x_2$ instead. So for your example, instead of an input of $18+24=42$ you will have to use an input of $18+2*24=66$ to get your answer of $15.5$.

Jaap Scherphuis
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COMMENT.- It is not possible because you have TWO DISTINCT functions. However if you remember the theory of conics it is a fact that there are cases in which the conic is degenerate to two right lines.

Because of you have two distinct lines $$ 32-x=4y\\24-x=2y$$ if you make the product $$(32-x-4y)(24-x-2y)=0$$ you get a degenerate conic whose graph is exactly the graph of the two lines above.

Piquito
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If I am right, your hope is to obtain a linear function $$y=ax+b$$

from two other linear functions

$$y=a'x+b'$$ and$$y=a''x+b''$$ in such a way that the image of the sum of two $x$'s is the sum of the corresponding $y$'s:

$$a(x_1+x_2)+b=a'x_1+b'+a''x_2+b''.$$

For this to occur, you need to identify the coefficients and ensure

$$\begin{cases}a=a'\\a=a'',\\b=b'+b''.\end{cases}$$

Then if $a'\ne a''$, this is impossible.

The best that you can do is to realize the equality for two particular pairs of $x$ values, say $x_1,x_2$ and $x_3,x_4$. After solving a $2\times2$ system for $a$ and $b$, there will be an infinity of pairs $x_5,x_6$ such that

$$a(x_5+x_6)+b=a'x_5+a''x_6+b'+b'',$$ for instance by setting

$$x_6=-\frac{(a'-a)x_5+b'+b''-b}{a''-a}.$$