Question 1187373: Kier is x years old and his sister Nancy is x^2 yrs. old. In 7 years, Nancy will be 1 year older than twice Kier’s age at that time. Find their present ages.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x = kier's age now.
x^2 = nancy's age now.
in 7 years, nancy will be 1 year older than twice kier's age at that time.
equation for that is:
x^2 + 7 = 2 * (x + 7) + 1
simplify to get:
x^2 + 7 = 2x + 15
subtract 2x + 15 from both sides of the equation to get:
x^2 + 7 - 2x - 15 = 0
combine like terms to get:
x^2 - 2x - 8 = 0
factor this quadratic equation to get:
(x + 2) * (x - 4) = 0
solve for x to get:
x = -2 or x = 4
x can't be negative, so x = 4.
that's kier's age now.
nancy's age now is x^2 = 4^2 = 16
in 7 years kier will be 11 and nancy will be 23.
23 = 2 * 11 + 1 = 22 + 1 = 23.
that says that, in 7 years, nancy will be 1 year older than twice kier's age then.
requirements of the problem have been satisfied.
solution is that kier is 4 years old now and nancy is 16.
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