for your first example i assume you dropped the x early? i.e.

2x + 5 = 7
7 - 5 = 2x
2 = 2x
2/2 = 2x/2
1 = x

it's just that it was a bit unclear, normally you keep your algabraic terms in the equation at all times.

in any equation you can add, subtract, divide or multiply both sides by anything, but you must do it to both sides. with that in mind...


4 - x = -4 lets add 4 to each side

4-x + 4 = (-4) + 4

8-x = 0 the two fours on the right cancel to give 0

now add x to each side

8 - x + x = (+)x
8 = x

I don't want to confuse you, but i think "start from the right and work your way back." is a confusing approach. It's better to try to get the x term on it's own, then you should be able to solve.

I find it strange how your program solves that equation. I prefer to see it another way: you have your equation, on which you can perform operations, which lead to equivalent equations (i.e. an equation which yields the same solutions).

If you have
a = b
then you can add the same number to both sides (in this case, c):
a = b ===> a + c = b + c
Adding a negative number is the same as subtracting, so this means you can subtract the same value from both sides as well:
a = b ===> a - c = b - c

A simple example on this would be
x - 5 = 9 ==> (x - 5) + 5 = 9 + 5 ==> x = 14

The same can be done with multiplication (and multiplying by 1/x is the same as dividing, so you can also do it with division), but beware, multiplying (or dividing) by 0 is not allowed.
x/7 = 2 ==> (x/7) * 7 = 2 * 7 ==> x = 14


There are several ways you can solve your equation:
4 - x = -4 ==> (4 - x) - 4 = -4 - 4 ==> -x = -8 ==> -x * -1 = -8 * -1 ==> x = 8
4 - x = -4 ==> (4 - x) + x = -4 + x ==> 4 = -4 + x ==> 4 + 4 = (-4 + x) + 4 ==> 8 = x