why does adding two equations that intersect when graphed result

>>9132407
Ironically, the definition of plotting a graph with two functions

>>9132407
y=f(x)
y=g(x)
for some c, y=f(c)=g(c)
adding the above equations gives
2y=f(x)+g(x)
for x=c
2y=f(c)+g(c)=2f(c)
2y=2f(c)
y=f(c)=g(c)

>the intersection of the two added equations.
i really dont get wtf youre talking about

>>9132430
you graph two equations in the form Ax+By = C

Example: y + -x = 3 and y+-2x = 5

when graphed these two equations intersect at (-2,1)

if we add the two equations we get 2y + -3x = 8, this new equation also passes through the point (-2,1)

I'm wondering why this is always true for any two added equations.

File: 1469466916876.jpg (8KB, 244x206px) Image search: [Google]

8KB, 244x206px

>>9132455
thanks. i was interpreting equations as functions for some reason

>>9132424
A thread died for this

>>9132455
take two equations Ax + By + C = 0, Dx + Ey + F = 0.
If you add them you get (Ax + By + C) + (Dx + Ey + F) = 0.
Therefore if (x,y) satisfies both equations, then both brackets in the new equation are zero, so you get 0 + 0 = 0 and so (x,y) satisfies the new equation as well.

>>9132463
Suppose you have system of equations [eqn]\begin{cases}A x+B y=C\\
D x+E y=F\end{cases}[/eqn]that intersect at a point [math](x,y)=(\alpha,\beta)[/math]. Then [math]\begin{cases}A \alpha+B \beta=C\\
D \alpha+E \beta=F\end{cases}[/math]. Adding the equations gives [math](A+D)\alpha+(B+E)\beta=C+F[/math] for the same reason that adding 2=1+1 and 4=2+2 gives 6=1+1+2+2=2+4