My Birthday Problem (easy)?

Let x = your age, and let y = the age of the bike.

Seven years ago on your birthday, you turned x - 7 years old. And, your bike was y - 7 years old. Since, as you say, it was half as old as you,

x - 7 = 2(y - 7)

=>

x = 2y - 7 .......... Eq. 1

Today, the bicycle is 4/7 your age. So,

x = (7/4)y .......... Eq. 2

Equating Eqs. 1 and 2, we get:

2y - 7 = (7/4)y

=>

8y - 28 = 7y

=>

y = 28

Substituting this value into Eq. 1, we get:

x = 2(28) - 7

= 49

So, you are 49 years old. Happy birthday!

I wanted to try your harder math problem, but I haven't learned about irreducible integrals yet! This one is already probably hard enough.

Let x = your age

y = Unknown years (the "relative" point) before

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

Seven years ago from some y time, the bike was half as old as you. Subtract 7 from x because x is your current age.

Eq 1 -------> y = (1/2)(x-7)

y + 7= (4/7)x

Today, the bike is four sevenths your age, and today is 7 years later from the y.

Eq 2 --------> y = (4/7)x - 7

So:

(1/2)(x-7) = (4/7)x - 7

x = 49

I'll think about this more.

I probably got it wrong, but happy birthday to you! Have a good one!

Let your present age be 7x

Age of bicycle = 4x

7 years ago

your age = 7x -7

age of bicycle = 4x -7

7x -7 = 2( 4x- 7)

7x -7 = 8x -14

x = 7

ANSWER = 49

CHECK

Bicycle = 4x = 28 years

7 years ago

your age = 49-7 = 42

bicycle = 28-7 = 21

42 = 2X 21

14?

14 - 7 = 7

Seven years ago the bike was half as old as you are now?

your 49

i think 16