#how to solve by combination?
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like have you seen
y+x=5
3y+x=10
subtracting you get -2y=-5
the second equation is just (x+y)/2-3(x-y)/5=1 so set both equations equal to each other (since they are both equal to 1)
oh okay let me try that
after expanding our a bunch and collecting like terms, youll arrive at a wuadratic equation you can solve, or an equation for a circle..
huh where is that
elimination
but the other method is better
can you please explain to me how you got to that point?
that’s just a random example unrelated to your question
oh
yea i need to do it by elimination
thats what I meant, sorry my maths is in french and idk the english terms
no 
could you not do this
what should I do?
tried this but still looks sketchy
idk, but I just guessed x=y=1 and it checks out.
Notice that 2 and 5 are of a different parity, but for integers x and y then x+y and x-y are always of the same parity. This means (along with the fact that 3 doesn't divide 5) that (x+y)/2 and 3(x-y)/5 cannot both be whole numbers for integers x and y, unless one of x+y or x-y is 0.
If you let x+y = 0, then the second equation cannot be solved in integers. So you should let x-y = 0.
This gives x=y=1 which also satisfies the first equation
There are probably more solutions tho
yes
,w (2+31)^2-21*(21+1)-31*(41+31-1)
$$\frac{x+y}2-\frac{3(x-y)}5-1=0\implies 10 = -x + 11y$$ but idk how to implement this without substitution