Inverse operations

Before solving equations with unknowns, try the idea on some arithmetic — where you can see exactly what is happening.

Example 1. It is known that 14593 − 1459 + 145 − 14 + 88 = 13353. What is the value of 14593 − 1459 + 145 − 14? Do we need to work that long string out? No. We can get it by subtracting 88 from 13353.

Why is that allowed? Because addition and subtraction are inverse operations. Subtracting 88 from both sides removes the term 88 from the LHS and leaves exactly the expression we wanted: 14593 − 1459 + 145 − 14 + 88 − 88 = 13353 − 88.

Example 2. It is known that 23 × 41 × 11 × 8 × 7 = 5,80,888. What is 23 × 41 × 11 × 8? Using the fact that multiplication and division are inverse operations, simply divide 5,80,888 by 7. And that is the same as dividing BOTH sides by 7, which removes the factor 7 from the LHS.

Notice how the two examples differ. In the first, the unwanted piece was a TERM, joined on by addition — so it is removed by subtracting. In the second it was a FACTOR, joined on by multiplication — so it is removed by dividing.

The same works when the unwanted term is a negative one. To remove a term −(−67), add (−67) to both sides; the two cancel and the RHS becomes 7091 + (−67). And to remove a fractional factor like 8/9, divide both sides by 8/9, which is the same as multiplying both sides by 9/8.

So every unwanted piece has an undoing: a term is undone by its additive inverse, a factor by DIVISION. Apply that undoing to BOTH sides, and the equality survives while the clutter disappears. That is the whole method, and it is about to be pointed at an unknown.