ode_to_joy said:
'adding a non-variable to an expression' means a 'transition' to certain direction. Say y=2x+3. If we add 5 to the right side of equation, we are moving the line 5 units upwards.
But I don't understand the rationale of adding variable to the equation to find a intersection point, or solving a systematic equation by substitution or elimination.
The others have given you some algebra courses to try.
I noticed that the way you wrote suggests you think of math operations as movements in space (or that this is a way that helps you?) I thought I'd try a description that used this:
First, I'll recap what you said, make sure I understand you,
then I will add a variable to an equation and show you what this does to it,
then I'll expand this idea to cover adding whole expressions together in different combinations.
Set:
y=2x+3
Adding a constant:
If we transform: y'=y+5
then y'=2x+8
...which is the original line translated 5 units in the y direction.
Adding a variable:
If we transform: y'=y+x
then y'=3x+5
...which is the original line rotated clockwise a bit.
We can also add two equations to each other:
(y=2x+3)+(y=4x+5)-->(2y=6x+8)-->(y=3x+4)
This rotates the lines
about the point that they intersect.
(sketch all three and you'll see it.)
This particular result is not terribly helpful when it comes to finding the intersection.
In general you can add multiples like this:
a(y=4x+5) + b(y=2x+3)
... to produce an arbitrary rotation about the intersection.
To find the intersection, what we want to do is rotate the line about the intersection until it is horizontal, so the equation is y=1 (in this case) and so you know the y-position of the intersection. From there finding the line x=-1 (in this case) is trivial.
Sounds rough but there is a trick to it. You learn the trick.
What you are probably being taught at High School is a collection of tricks for different classes of simultaneous equations. For the example, if you put a=1 and b=-2 this happens:
1(y=4x+5) + -2(y=2x+3) = (-y=0x-1) so (y=1) see?
(again, a sketch will show you what is happening)
This is "elimination" of x: it rotates the line about the intersection so it is horizontal (eliminate y and you get a vertical line.)
Of course, you could do y'=y-4x in the first equation - that will give you a horizontal line too. Unfortunately it did not rotate about the intersection so y=5 (means x=0) will not be the correct intersection.Thinking of these operations as movements in space is nice going, but will trip you up when you deal with more abstract math. Best practice is to get used to different ways of thinking about math.
How someone would normally do the last example, is to realize that y has to have the same value in both equations and reason that the only way that happens is if
(4x+5)=(2x+3)
... (just putting y=y) so you write that down then use the rules of algebra to make x the subject (called "solving for x"). This way looks for the vertical line first. (Eliminating y, by putting a=1 and b=-1 in the first method.)
It is equally valid to make x the subject of both equations, and put x=x:
(y-5)/4 = (y-3)/2
Also equally valid to make x the subject of one, then substitute it into the other, then solve for y. Which sounds more complicated but is usually easier to do:
x=(y-5)/4 ...(from the 1st equation, sub into the second)
y=2(y-5)/4 +3
These are all doing the same things.
Sometimes it is not immediately obvious which a and b to use when adding equations (first method) together - so we break the process down into easier steps ... which is what the last three methods were doing. We pick the method that makes the math easier for us.