Math, asked by dheerajgawli1971, 3 months ago

2 years ago my son age was 4 1 upon 2 times the age of my son than six years ago my age was twice the square of the Age Of My son and then find the present age of my son​

Answers

Answered by palaksonisoni955
0

Answer:

10 years

Step-by-step explanation:

let the present age of me be x and my son be y

two years ago ,

my age was x-2 and

my son's age was y-2

it is given that 2 years ago my age was 4 and half times times of the age of my son . or in other words,my age was two times the age of my son.

so,(x-1)=4.5(y-2)

= x=(9/2)y-7 ...1)

6 years ago, my age was x-6 and my son age was y-6 .

it is given that my ageaged twice the square of the age of my son .

so,(x-6)=2×(y-6)²

substituting the value of x from (1),we get ;

(9/2)y- 7-6= 2(y-6)²

=(9/2)y- 13 =2(y-6)²

=(9/2)y-13= 2y²+72-24y

=(9/2)y =2y²-24y +85

=9y= 4y²-48y +170

=4y²-57y+170=0

=4y²-40y-17y+170 =0

=4y (y-10) -17(y-10)=0

=(y-10) (4y-17)

by zero product rule ,

We get y-10 and y=17/4=4.25 y=4.25 is discarded as age cannot be fractional

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