1.John’s age is three times that of his sister. When you add John’s age to his sister's age, you get 36. How old are John and his sister? (a)Write an equation that represents the situation. (1 point)
(b)Solve the equation from Part (a). Show your work. (3 points)


Part A: Write an equation.
If the present age is y, then n times the present age = ny.
If the present age is x, then age n years later/hence = x + n.
If the present age is x, then age n years ago = x – n.
The ages in a ratio a: b will be ax and bx.

Part B: Solve the equation.

John’s age =

John’s sister’s age =

sister = 6 years

John=18 years

Step-by-step explanation:

A(S) + 12 = 3 * A(S), subtracting A(S) from both sides one gets 12 = 2 * A(S). Now dividing both sides by 2 we get A(S) = 12/2 = 6 . Therefore, the age of the sister is 6 years when her brother John is three times as old, that is 18 years.

36-x=y

y*3=x

x+y=36

Step-by-step explanation:

I wasnt really sure how to write the equation the way you're supposesd to?

John is 27 and his sister is 9. 9*3=27

I just used guess and check sorry

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.