Question 979008: Lucas is 9 years older than three times his son’s age. In 14 years, the age of Lucas will be 6 years more than twice his son’s age. Find their present ages.
PARTIAL SOLUTION
x= son's age. So lucas age= 3x+9
After 14 years, son age= x +14, lucas age= 2(x+14)+6
Not sure how to move ahead
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39616)   (Show Source):
Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website! Lucas is 9 years older than three times his son’s age. In 14 years, the age of Lucas will be 6 years more than twice his son’s age. Find their present ages.
PARTIAL SOLUTION
x= son's age. So lucas age= 3x+9
After 14 years, son age= x +14, lucas age= 2(x+14)+6
Not sure how to move ahead
After 14 years, the son will be x + 14 -------- You're correct up to here
After 14 years, Lucas' age will be: 3x + 9 + 14, or 3x + 23
Since at that time (14 years hence), Lucas' age will be 6 years more than twice his son's age (the son's age in 14 years, I presume), then we can say that:
3x + 23 = 2(x + 14) + 6
3x + 23 = 2x + 28 + 6
3x + 23 = 2x + 34
3x - 2x = 34 - 23
x, or son's current age: 
Lucas' current age: 3(11) + 9, or 33 + 9, or 
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