If $p+q+r=70, p=2 q$, and $q=3 r$, what is the value of $p$ ?
Solution 1

Substituting $p=2 q$ into $p+q+r=70$ gives $2 q+q+r=70$ or $3 q+r=70$.

Substituting $q=3 r$ into $3 q+r=70$ gives $3(3 r)+r=70$ or $10 r=70$ which means $r=7$.

Substituting $r=7$ into $q=3 r$ gives $q=21$, and substituting $q=21$ into $p=2 q$ gives $p=42$.

## Solution 2

The equation $p=2 q$ is equivalent to $q=\frac{1}{2} p$.

The equation $q=3 r$ is equivalent to $r=\frac{1}{3} q$.

Substituting $q=\frac{1}{2} p$ into $r=\frac{1}{3} q$ gives $r=\frac{1}{6} p$.

Substituting these values of $q$ and $r$ into $p+q+r=70$ gives $p+\frac{1}{2} p+\frac{1}{6} p=70$.

Therefore,

$$
p=\frac{70}{1+\frac{1}{2}+\frac{1}{6}}=\frac{70}{\frac{10}{6}}=42
$$

ANSWER: 42