I guess I'm lucky. Half of my students are "regular" students, but I don't have too many problems. I do work with some small groups of students who need extra help though, and almost invariably, the kids who are struggling, lack basic arithmetic skills. It's pretty hard to figure out what's going on with 8x times 9x = 72x², when you don't even know that 8 times 9 = 72. "It's like, whoa, magic, everything changed."
My only advice when working with kids like this is a very step by step process that relies 100% on their calculators. Adding and subtracting vertically helps. (And expecially show all the work, like on the first problem you linked.)
When solving a problem like 5x+3=7x-9, it pains me to have the students automatically subtract the 7x from both sides. But, when I work with remedial level students, that's exactly what we do. Most people would just mentally arrange it, while avoiding negative numbers, but with remedial students, it really doesn't matter - they're going to be using the calculator for that.
There are all sorts of "crutches" you can give to remedial level students to make it so that they can at least pass high school algebra. Percent problems become "is over of equals percent over 100." Again, I cringe. But, from a more experienced teacher, I learned that even the worst students get the majority of simple problems correct when writing that proportion and then filling in the numbers.
In geometry, other than doing proofs, I've never had trouble with any level of students. But, you have to be aware that many of the lower level students are pretty poor at memorizing formulas such as finding the midpoint. With the slope formula, distance/length of a line segment formula, and midpoint formula, there are so many x+x's and x-x's that the students often confuse them (the formulas.) So, midpoint becomes "average the x's, average the y's."
Other types of problems simply become a matter of drill and kill. I had a student last year who couldn't do a proof to save his/her life. "AB is perpendicular to BC. What can you figure out from that?" "That angle A is congruent to angle C because the perpendicular transversal bisects the midpoint of the pythagorean isosceles triangle?" <bewildered look on my face.>
But you can damn well bet that he/she, given two points, could find the equation of the perpendicular bisector of the segment connecting those two points. Blindfolded. In his/her sleep. Drill & kill baby. What's nice about that is when you can find a "difficult" problem like that and FINALLY get them to accomplish one independently, it really boosts their self esteem in mathematics a little bit. Then, you can play off that boost & get them to learn a little more. But, at some point, you have to figure out where to cut your losses (in geometry, it's usually proofs) and more effectively utilize the remaining time.
