1. Each of Alice and Beatrice has their birthday on the same day.
In 8 years' time, Alice will be twice as old as Beatrice. Ten years ago, the sum of their ages was 21 .
How old is Alice now?
Solution
Let Alice be $a$ years old now and let Beatrice be $b$ years old now.
In 8 years' time, Alice will be twice as old as Beatrice, so that
$$
a+8=2(b+8),
$$
which we may simplify to
$$
a-2 b=8 .
$$

Ten years ago, the sum of Alice's age and Beatrice's age was 21, so that
$$
a-10+b-10=21,
$$
which we may rewrite in the form
$$
a+b=41 .
$$

Adding equation (1) to $2 \times$ equation (2) in order to eliminate $b$, we get $3 a=90$, so that $a=30$. Hence Alice is now 30 years old.