Using formula: $P_1 = -20$, $P_2 = 25$
$P_{net} = -20 + 25 + \frac{(-20) \times 25}{100} = 5 + \frac{-500}{100} = 5 - 5 = 0\%$
No net change.

For $n$ successive changes ($P_1, P_2, ..., P_n$), the factor approach is generally easier:

$NCF = (1 + \frac{P_1}{100}) \times (1 + \frac{P_2}{100}) \times ... \times (1 + \frac{P_n}{100})$

$P_{net} = (NCF - 1) \times 100\%$

Problem: A population increases by $5\%$, then by $8\%$, and finally decreases by $10\%$. What is the net percentage change in population?
Using factors: $P_1 = 5, P_2 = 8, P_3 = -10$
$NCF = (1 + \frac{5}{100}) \times (1 + \frac{8}{100}) \times (1 - \frac{10}{100})$
$NCF = 1.05 \times 1.08 \times 0.90$
$NCF = 1.134 \times 0.90 = 1.0206$
$P_{net} = (1.0206 - 1) \times 100 = 0.0206 \times 100 = 2.06\%$
Net increase of $2.06\%$.
Using two-change formula iteratively:
First two changes ($5\%$ and $8\%$):
$P_{net12} = 5 + 8 + \frac{5 \times 8}{100} = 13 + \frac{40}{100} = 13 + 0.4 = 13.4\%$
Now combine $13.4\%$ with the third change ($-10\%$):
$P_{net} = 13.4 + (-10) + \frac{13.4 \times (-10)}{100} = 3.4 + \frac{-134}{100} = 3.4 - 1.34 = 2.06\%$
Same result.

When multiple discounts are applied, they are successive percentage decreases.
Problem: A shirt is discounted by $20\%$ and then an additional $10\%$ off the discounted price. What is the single equivalent discount?
Using formula: $P_1 = -20, P_2 = -10$
$P_{net} = -20 + (-10) + \frac{(-20) \times (-10)}{100} = -30 + \frac{200}{100} = -30 + 2 = -28\%$
Equivalent to a single $28\%$ discount. (Not $30\%$)

If the sides of a 2D shape change by percentages, the area changes by a net percentage.
Problem: The length of a rectangle increases by $20\%$ and its width decreases by $10\%$. What is the net percentage change in its area?
Using formula: $P_L = 20, P_W = -10$
$P_{Area} = 20 + (-10) + \frac{20 \times (-10)}{100} = 10 + \frac{-200}{100} = 10 - 2 = 8\%$
The area increases by $8\%$.

The order of percentage changes does NOT affect the net percentage change.
$P_1 + P_2 + \frac{P_1 P_2}{100} = P_2 + P_1 + \frac{P_2 P_1}{100}$
Example 3 and 4 showed this: Increasing by $25\%$ then decreasing by $20\%$ gives $0\%$ change, and decreasing by $20\%$ then increasing by $25\%$ also gives $0\%$ change.

If a final value $Q_f$ is known after a net percentage change $P_{net}$, and you need to find the original value $Q$.
$Q_f = Q \times (1 + \frac{P_{net}}{100})$
$Q = \frac{Q_f}{(1 + \frac{P_{net}}{100})}$
Problem: After a $20\%$ increase, a product costs $ \$120$. What was its original price?
$120 = Q \times (1 + \frac{20}{100}) = Q \times 1.20$
$Q = \frac{120}{1.20} = \$100$

If the net percentage change and one of the successive changes are known, you can find the other.
Problem: A price increased by $10\%$, and the net increase after a second change was $21\%$. What was the second percentage change?
Let $P_1 = 10$, $P_{net} = 21$. Find $P_2$.
$21 = 10 + P_2 + \frac{10 \times P_2}{100}$
$21