sanjay1810 wrote:

hello,

i understood Bunuel's solution. However, i'd a question: how does one decide if one should use back substitution or equations to solve? in this case, i tried solving with equations. since we don't know sign of a, i solved for both and got 2 and 8 as answers then back substituted to check ( since both 2 and 8 are in answer choices) and then found only 9 works. any advise?

thanks.

If you take a step back and look at the operation it's pretty simple. If x > 0, subtract 1. If x < 0, add one. The initial number then definitely can't be negative or lower than 5 since you'd then have no way to get up to 5. Hence, if you want to get to 5 you have to start higher, since eventually you'll just keep getting alternating values of 0, -1, 0, -1, 0, -1.

If the operation isn't that simple it's a game time decision I'd think. I typically just backsolve. The only time I'd consider not backsolving is if the answer options are nasty or if the operation is "normal" and just a function - f(k) or f(x) or whatever.