Question 795284: 1.) Alice is twice as old as Jimmy. In Twelve years, the sum of their ages is 99. How old is each now?
2.) A certain doctor's age is no 7/9 of what it will be 18 years from now. Just how old is he now?
3.) The sum of the ages of a father and her daughter is 44 years old. The father's age is 4 years more than 3 times the daughter's age. find the ages of both.
Answer by wilft1(217)   (Show Source):
You can put this solution on YOUR website! first one
x + 2x + 12 + 12 = 99 (in twelve years the sum of their ages is 99, there are two of them, thats why i put in 12 + 12)
combine like terms...
3x + 24 = 99
subtract 24 from both sides
3x = 75
divide both sides by 3
x = 25
jimmy is 25, alice is 50
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doc is 7/9 of what he will be 18 years from now. that tells me that 18 is 2/9 of his age in the future, so multiply 18 x 4.5 = 81 (4.5 is how many times 2 will go into 9 and make the fraction a whole number). you can make it a little easier by saying that it will be 9 of 1/9, so just multiply 9 x 9 = 81. 18 years from now he will be 81, so subtract 18 from 81 to learn that his current age is 63.
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father is 4 years more than 3 times his daughters age, equals out to 44 years old... the formula for this...
x + 3x + 4 = 44
subtract 4 from both sides...
x + 3x = 40
combine like terms
4x = 40
divide both sides by 4
x = 10
daughter is 10 years old, daddy is 30 years old
hope this helps :)
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