How to simplify expression depending on former definition?

Question

If

a = x + y
b = x - y
u = a + b
v = a

This gives me for u/v the expression (2 x)/(x + y) But I want to get 1+b/a. How to do that when a and b are very long expressions?

Why does it matter if a and b are long expressions if you just want 1+b/a? — lericr, Dec 21 '23 at 06:19
1+b/a is ok, I just wanted to avoid long Replace answers. I was looking for some kind of short switch (from x,y to a,b), because I have many variables to deal with. I was looking sore something like Eliminate, but that works only for equations. — user57467, Dec 21 '23 at 07:21
Does this answer your question? Can I simplify an expression into form which uses my own definitions? — MarcoB, Dec 21 '23 at 12:31
I found that the following works eq1 = a == x + y; eq2 = b == x - y; eq3 = u == a + b; eq4 = v == a; eq5 = c == u/v; sol = Eliminate[{eq1, eq2, eq3, eq4, eq5}, {x, y, u, v}]; Solve[sol, c] Which looks very clumsy. Any better ideas? — user57467, Dec 21 '23 at 13:44
What are you going to do with the 1+b/a expression? If you just want to see it, you can use one of the Hold forms (for example). — lericr, Dec 21 '23 at 16:15
Lericr, How would I do it then with Hold forms? I want to use different forms by which you can describe triangles (by a,b,c or angles or other parts once coordinates are given) — user57467, Dec 22 '23 at 12:31

score 3 · Answer 1 · answered Dec 21 '23 at 05:59

3

u/v /. u -> a + b /. v -> a // Apart

answered Dec 21 '23 at 05:59

bmf

1 Answers1