Question

Two years ago my age was 9/2 times my son's age.six years ago my age was twice the square of my sons age . Find my sons present age.solve using quadratic equation

Answer 1

Here is your answer

HOPE IT IS USEFUL

Answer 2

My age is 38 years

Son's age is 10

Step-by-step explanation:

Let the Son's age be x

Let my age be y

Son's age 2 years ago = x-2

My age 2 years ago = y-2

We are given that Two years ago my age was 9/2 times my son's age

$y-2=\frac{9}{2}(x-2)\\2y-4=9x-18$

9x-2y=14 ---- 1

Son's age 6 years ago = x-6

My age 6 years ago = y-6

Six years ago my age was twice the square of my sons age

$y-2=\frac{9}{2}(x-2)\\2y-4=9x-18$

$y-6=2x^2+72-24x$ ------ 2

Plot the lines on the graph

9x-2y=14 --- Purple line

$y-6=2x^2+72-24x$   ---- Black line

(x,y)=(10,38)

My age is 38 years

Son's age is 10

#Learn more:

Eight years ago a ashwin age is one less than three times that of arpit age.six years ago ashwin age was one more than 2 times that of arpit age.what is arpits age

https://brainly.in/question/10397748