.
"p" notebooks costed N200 ====> hence, one notebook price was before increasing.
After increasing, one notebook price becomes , and the difference is N10.
Equation
- = 10
To solve it, multiply both sides by p*(p-1). You will get
200*p - 200*(p-1) = 10p*(p-1)
200p - 200p + 200 = 10p^2 - 10p
10p^2 - 10p - 200 = 0
p^2 - p - 20 = 0
(p-5)*(p+4) = 0.
The only positive solution is p= 5.
Answer. Before price increasing, one notebook cost was = N40.
Solved.
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To see many other similar problems solved by the same method, look into the lesson
- Challenging word problems solved using quadratic equations
in this site.