Question

the sum of the ages of brother and sister is 20.4 yrs ago the product of their ages in years was 32 . is it possible to determine their present ages

Answer 1

it is possible to find there present ages with this information
if you will solve it it will as follows
age of brother = x
age of sister = y
so x+y=20.....(2)
so y = 20-x.....(1)
there age before four years
(x-4) & (y-4)
there product is 32 (given)
so
(x-4) x (y-4) = 32
by eqn 1
(x-4) x (20-x-4) = 32
(x-4) x (16-x) = 32
multiple -1 both side
(x-4) x (-(16-x)) = -32
(x-4) x (-16+x) = -32
(x-4) x (x-16) = -32
x^2 -16x -4x +64= -32
x^2 - 20x +96=0
x^2 -12x -8x +96=0
x(x-12)-8(x-12)=0
(x-8)(x-12)=0
(x-8)=0/(x-12) or. (x-12)=0/(x-8)
x-8=0. or. x-12=0
x=8. or. x= 12

if x ( brother' age ) = 8
then y ( sister' age ) = 20-8=12( byeqn 2)

if x ( brother' age ) = 12
then y ( sister' age ) = 20-12=8( byeqn 2)

there are two answer to this question

Answer 2

let the age of brother =x
let age of sister =y
therefore ....x+y=20
x=20-y
four years ago,
age of sister=y-4 and age of brother =x-4
therefore.....y-4*x-4=32
y-4*20-y-4=32
y-4*16-y=32