Five years ago, a mother was twice as old as her son. In 6 years' time, the sum of their ages will be 82. Find their present ages

Question

It is a word problems leading to linear equations

Answer 1

Hello, Princess!

We are going to set up two equations. One for the mother's and son's ages 5 years ago, and another for their ages in 6 years.

The easiest way to do this is to make the first equation be our baseline of sorts, in other words, rather than saying 5 years ago, we will simply say that m=2s (mother's age = 2 times son's age). So, our first equation is m=2s.

The second equation we will set up as this: m+11+s+11=82. We are saying the mother's age plus 11 years, plus the son's age plus 11 years (because m=2s is for 2009), is equal to 82.

Now, we have these two equations: m=2s, and m+11+s+11=82.

If we solve the second equation for m, we get m=82-11-11-s, or m=60-s

The reason we solved the second equation for m, is because we can now plug in this m's value into the first equation to solve for s, or the son's age.

Thus: 60-s=2s. Add s to both sides to get 60=3s, then divide both sides by 3 to get s=20.

So, five years ago, the son was 20, and the mother was 40 (twice his age).

We can be sure of this because if we plug those ages back into the second equation, 40+11+20+11, we get 82.

We're not done yet, though. Remember the ages we just garnered were 5 years ago. So, for the present ages, simply add 5 to each. Therefore, the mother's age is 45, and the son's is 25.

Hope this helped!