Question

The sum of the present ages of a brother and his younger sister is 16 years. If the product of their age is 63, find their ages.​

Answer 1

Let the age of brother be x and that of a sister be y.

(x+y) =16

(x) ×(y) =63

Using the substitution method, x=16-y

Substitute in equation 2

(16-y)(y)=63

16y-y^2=63

Y^2-16y+63=0

P=63

Sum= -16

Factor=(y-7),(y-9)

(y-7)(y-9)=0

Therefore y=7 or y=9

Therefore age of brother = 16-7 =9 0r 16-9=7

Therefore their ages are 9 years and 7 years