Trial and error, and the balance property

How did you solve the equations you framed? One way is to SUBSTITUTE different values in place of the letter-number and check which one makes LHS = RHS.

Try it on 2n + 1 = 99. Put n = 5: the LHS is 2 × (5) + 1 = 11, far from 99. Put n = 10: the LHS is 21, still not equal. Put n = 30: the LHS is 61, still much lower.

Put n = 40: the LHS is 81, getting closer. Put n = 50: the LHS is 101, just a bit too high. Put n = 49: the LHS is 99. So the solution to 2n + 1 = 99 is n = 49. (Could this equation have any OTHER solution?)

This is called the trial and error method. It works, and it took six guesses. Try solving 5x − 4 = 7 the same way and you will feel the problem: the trial and error method can be INEFFICIENT, and it is worse when the answer is not a whole number.

Recall that with the weighing scale we did NOT use trial and error. For some problems the solution was straightforward; for others we used the fact that removing equal weights from both plates keeps the scale balanced. Do equations have a similar property?

Test the idea on an equation with no unknown at all: 15 + 8 = 23. If we add, subtract, multiply or divide by the same number on both sides, will the equality survive? Check by adding 10 to both sides: 15 + 8 + 10 = 33 and 23 + 10 = 33. It does.

And the reason is plain. Since the LHS and the RHS of an equation have the same value, performing the same operation on both of them clearly cannot change their equality. That one sentence replaces guessing with a method.