Asked by Sylvestre
Aris is twice as old as Rico while Jay is 24 years younger than Aris. If half of Aris' age six years ago was three less than one half the sum of Rico's age in four years and Jay's present age, then what is their ages before and now.
Answers
Answered by
Jai
Let A = Aris' age
Let R = Rico's age
Let J = Jay's age
From the problem,
Aris is twice as old as Rico:
A = 2R
Jay is 24 years younger than Aris:
J = A - 24
half of Aris' age six years ago was three less than one half the sum of Rico's age in four years and Jay's present age
(1/2)(A - 6) = (1/2)(R + 4 + J) - 3
You have an equation for A and J (the first two equations above). You can substitute it to the third equation to solve for R. After solving for R, solve for the rest.
Hope this helps~ `u`
Let R = Rico's age
Let J = Jay's age
From the problem,
Aris is twice as old as Rico:
A = 2R
Jay is 24 years younger than Aris:
J = A - 24
half of Aris' age six years ago was three less than one half the sum of Rico's age in four years and Jay's present age
(1/2)(A - 6) = (1/2)(R + 4 + J) - 3
You have an equation for A and J (the first two equations above). You can substitute it to the third equation to solve for R. After solving for R, solve for the rest.
Hope this helps~ `u`
Answered by
Anonymous
45
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