# The square of Janets age is 400 more then the square of the sum of Kim's and Sue's ages. Kim's and Sue's ages total 10 less then Janets age. Find the square of the ages of Janet, Kim, and Sue

Let Janet's age to be `j` , Kim's age to be `k` , and Sue's age to be `s` .

The square of Janet's age is 400 more than the square of the sum of Kim's and Sue's ages. This information can be represented algebraically as:

`j^{2} = (k+s)^{2} + 400`

Kim's and Sue's ages total 10 less than Janet's age, that is:

`(k+s) + 10 = j`

It then means that `j^{2}` is also `((k+s) + 10)^{2}` from the second equation

We substitute this value of `j^{2}` in the first equation to give

`(k+s)^{2} + 20(k+s) + 100 = (k+s)^{2} + 400`

The `(k+s)^{2}` in both sides of the equation cancel out to yield

`20(k+s) = 300`

So that `(k+s) = 15`

and `j = 15 + 10 = 25`

The square of Janet's age is `(25)^{2} = 625`

However, we need further information in order to calculate Kim and Sue's ages. The square of the sum of Kim and Sue's age is `(k +s)^{2} = 15^{2} = 225`

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Let Janet's age be "x".

Kim+Sue's ages = x - 10

The square of Janet is `x^2.`

The square of the sum of Kim & Sue's ages = `(x-10)^2`

Since Janet's is 400 more,

`x^2 = (x-10)^2 + 400`

`x^2 = x^2-20x+100+400`

`20x = 500`

`x = 25`

Janets is 25 years old, and Kim and Sue's ages together total 15.

The square of Kim's age is 625.  The square of the sum of Kim and Sues ages is 225.

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