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The square of Janets age is 400 more then the square of the sum of Kim's and Sue's...

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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

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Posted (Answer #1)

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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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primrose1e's profile pic

Posted (Reply #1)

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thank you so much

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Posted (Answer #2)

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The square of Janet's 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 Janet's age.

If Janet, Kim and Sue's ages are represented by variables J, K and S we have:

J^2 = 400 + (K + S)^2

K + S = J - 10

Combining the two equations J^2 = 400 + (J - 10)^2

J^2 - (J - 10)^2 = 400

(J - J + 10)(J + J - 10) = 400

10*(2J - 10) = 400

2J - 10 = 40

2J = 50

J = 25

K + S = 15

This gives the square of Janet's age as 625 and the square of the sum of Kim and Sue's age as 225. It is not possible to determine the age of Kim and Sue individually.

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