Two years ago Charles was three times her son's age and in 11 years time she will be twice as old. Find their present ages. Find out how old they are now?
1 Answer
OK, firstly we need to translate the words into algebra. Then we’ll see if we can find a solution.
Explanation:
Let’s call Charlie’s age, c and her son’s, s
The first sentence tells us c - 2 = 3 x s (Eqn 1j
The second tells us that c + 11 = 2 x s (Eqn 2)
OK, now we have 2 simultaneous equations we can try to solve them. There are two (very similar) techniques, elimination and substitution, to solve simultaneous equations. Both work, it’s a matter of which is easier. I’ll go with substitution (I think that was the category you posted it in.)
Let’s rearrange equation 1 to give: c = 3s + 2 (Eqn 3)
Now we can put that value for c back into equation 2 (this is the substitution bit)
Substituting from Eqn 3 into Eqn 2 gives: (3s + 2) + 11 = 2s (Eqn 4)
Simplifying, we put all the ‘s’ terms on one side (-2s from both sides) and collect all the digits on the other side, gives us:
s = -13 which is odd.
Children normally have a positive age. This would suggest (from Eqn 1) that Charlie’s age is 41 as c - 2 (39) is 3s. That works ok.
