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The operation rules are as follows：The coefficients in the equation are not equal to zero，That is to say, in the process of converting to an integral equation, if a common factor is proposed and the right side of the equal sign is zero, or if there is the same common factor on both sides of the equation, it can be reduced

Assume that the equation is：

a^2 y0 -a^2 y0 (y y0)/b^2 ==x y0 x0

The desired result of the integral equation is:

b^2 x x0 + a^2 (-b^2 + y y0)==0

In this example, x0 is not equal to 0, how to transform this fractional equation into an integral equation. (5 a + 8 c + d + 15 a x0 + 24 c x0 + 3 d x0)/x0 == 0

fractional equation

(5 a + 8 c + d + 15 a x0 + 24 c x0 + 3 d x0)/x0 == 0,

x0 is not equal to 0.transform this fractional equation into an integral equation.

Because x0 is not equal to 0, the left and right sides of the equation can be multiplied by x0 at the same time.

Get：

5 a + 8 c + d + 15 a x0 + 24 c x0 + 3 d x0==0，

5 a + 8 c + d + 15 a x0 + 24 c x0 + 3 d x0==0

Can be decomposed by factor ，the result is

(3 x0 + 1) (5 a + 8 c + d)==0，

x0 is not equal to 0.

So the end result is

5 a + 8 c + d==0

This is the logic of calculation.

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  • $\begingroup$ Could you add some practical example with operators and functions to understand your logic? $\endgroup$
    – Alex Trounev
    Jan 27 at 6:05
  • $\begingroup$ fractional equation(5 a + 8 c + d + 15 a x0 + 24 c x0 + 3 d x0)/x0 == 0,x0 is not equal to 0.transform this fractional equation into an integral equation. Because x0 is not equal to 0, the left and right sides of the equation can be multiplied by x0 at the same time.Get5 a + 8 c + d + 15 a x0 + 24 c x0 + 3 d x0==0，t5 a + 8 c + d + 15 a x0 + 24 c x0 + 3 d x0==0Can be decomposed by factor ，the result is (3 x0 + 1) (5 a + 8 c + d)==0，x0 is not equal to 0.So the end result is5 a + 8 c + d==0 $\endgroup$
    – csn899
    Jan 27 at 6:33

1 Answer 1

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$\begingroup$ 
eq = a^2 y0 - a^2 y0 (y y0)/b^2 == x y0 x0;
Assuming[y0 != 0, DivideSides[Factor@SubtractSides[eq], y0/b^2]]

mma

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  • $\begingroup$ thank you! Does this method first specify that there is a common factor y0 on the left and right sides of the equation? $\endgroup$
    – csn899
    Jan 27 at 3:06
  • $\begingroup$ @csn899 as you write the command in one line, it is very intuitive to follow each command and check what it does. If you follow closely, Factor@SubtractSides[eq] returns precisely the y0/b^2 factor which I used to DivideSides. So, to answer your question, yes this approach tells you how you factorized the expression, and the term you need to divide both sides. $\endgroup$
    – bmf
    Jan 27 at 3:10
  • $\begingroup$ Can you automatically find out the common factors on the left and right sides of the equation and simplify them into integral equations? $\endgroup$
    – csn899
    Jan 27 at 3:37
  • $\begingroup$ @csn899 what I wrote above should work in general. If your question is how to fill in the x in DivideSides[stuff,x] in general, my answer to you is I don't know how to do it, but I don't think it can be done in all generality. The reason being that different equations, factorize differently and different things have to be non-zero in order for the operation to make sense. It's not too cumbersome to apply the above in other examples btw $\endgroup$
    – bmf
    Jan 27 at 3:43

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