six years before the age of mother was equal to the square of her sons age. Three year hence her age will be thrice the age of her son then find the present ages of the mother and son
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let age of mother and sun be x,y respectively
Six years before
x-6 = ( y-6)^2 .....(1)
Three year hence
x +3 = 3( y+3)
x+3 =3y +9
x-3y= 6.
x= 6+3y
put in 1)
x-6 = ( y-6)^2
6 +3y -6 = ( y-6)^2
3y= y^2 +36 -12y
y^2 -15 y +36= 0
y^2 -12y -3y +36= 0
y( y-12) -3(y-12) = 0
y= 3,12
x= 6 +3y = 6+3(3)= 6+9= 15
x= 6+3(12) = 6 +36= 42
Six years before
x-6 = ( y-6)^2 .....(1)
Three year hence
x +3 = 3( y+3)
x+3 =3y +9
x-3y= 6.
x= 6+3y
put in 1)
x-6 = ( y-6)^2
6 +3y -6 = ( y-6)^2
3y= y^2 +36 -12y
y^2 -15 y +36= 0
y^2 -12y -3y +36= 0
y( y-12) -3(y-12) = 0
y= 3,12
x= 6 +3y = 6+3(3)= 6+9= 15
x= 6+3(12) = 6 +36= 42
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