# How to solve age word problems?

Roy is now 4 years older than Erik and half of that amount older than Iris. If in 2 years, roy will be twice as old as Erik, then in 2 years what would be Roy's age multiplied by Iris's age?

Is there any general method or are they all so confusing?

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You can try forming some systems of linear equations to solve them. – user130512 Mar 7 '14 at 20:23
Write the associated equation. – jinawee Mar 7 '14 at 20:23
that's the way to go^^ – Max Mar 7 '14 at 20:24

Roy is now 4 years older than Erik

$$R = E + 4$$

and half of that amount older than Iris.

$$R = I + 2$$

If in 2 years, roy will be twice as old as Erik

$$(R + 2) = 2(E + 2)$$

then in 2 years what would be Roy's age multiplied by Iris's age?

$$(R + 2)(I+2) = \;\; ?$$

Then you just solve the equations. Hopefully you can see that this method will apply to any similar problem equally well.

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This cleared the concept for me! Thanks! – Prakash Wadhwani Mar 8 '14 at 5:02
The answers is 48! – Prakash Wadhwani Mar 8 '14 at 5:04

Hint: Set $E$ the age of Erik, $I$ the age of Iris and $R$ the age of Roy today. You are given that $$R=4+E \tag{1}$$ and $$R=2+I \tag{2}$$ if I understood correctly this one (I am not sure what "that amount" is in your formulation) and $$(R+2)=2(E+2) \tag{3}$$ and you want to find $$(R+2)\times (I+2)$$ (if I understood that correctly too). Now you can solve the system of the three equations (1), (2), (3) (please check if they are formulated correctly) in three unknowns $E, I$ and $R$. That is the general method, with the difficulty being in the correct formulation of the given relations.

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I think the second equation should be $R=I+2$ – user130512 Mar 7 '14 at 20:27
@user130512 So with "this amount" is meant this 4? It is not clear – Jimmy R. Mar 7 '14 at 20:30
The question definitely should have been worded better, but I think that because this is a precalculus algebra question, they will not be asking him to solve for R as a function of both E and I – user130512 Mar 7 '14 at 20:32
@user130512 Ok, thanks, I corrected it. Seems more legitimate – Jimmy R. Mar 7 '14 at 20:34