7.6. Two people $A$ and $B$ need to travel from point $M$ to point $N$, which is 15 km away from $M$. On foot, they can move at a speed of 6 km/h. In addition, they have a bicycle that can be ridden at a speed of 15 km/h. $A$ sets out on foot, while $B$ rides the bicycle until meeting pedestrian $C$, who is walking from $N$ to $M$. After the meeting, $B$ continues on foot, and $C$ rides the bicycle until meeting $A$ and hands over the bicycle to him, who then rides it to $N$.

a) When should pedestrian $C$ leave from $N$ so that $A$ and $B$ arrive at point $N$ simultaneously (if he walks at the same speed as $A$ and $B$)?

b) When should pedestrian $C$ leave from $N$ so that the time spent by $A$ and $B$ on the journey to $N$ is minimized (if $C$ walks at the same speed as $A$ and $B$; the time spent on the journey is calculated from the moment $A$ and $B$ leave $M$ until the moment the last of them arrives in $N$)?

## 7.3. Inequalities
7.6. a) For $A$ and $B$ to arrive at point $N$ simultaneously, they must walk the same distance $x$ km and cycle the same distance $15-x$. Then $C$ must also walk the distance $x$ before meeting $B$. Therefore, $C$ will cycle $15-2x$ before meeting $A$, and after that, $A$ will cycle $15-x$. Together, $A$ and $C$ will cycle a total of $30-3x$. In the same amount of time, $B$ will walk $x$, so $\frac{30-3x}{15} = \frac{x}{6}$, i.e., $x = \frac{60}{11}$ km. $B$ will cycle $\frac{15-x}{15} = \frac{1}{15} \cdot \frac{105}{11} = \frac{7}{11}$ hours before meeting $C$, and $C$ will walk $\frac{x}{6} = \frac{1}{6} \cdot \frac{60}{11} = \frac{10}{11}$ hours before meeting $B$, so $C$ should leave $M$ $\frac{3}{11}$ hours before $A$ and $B$ start their journey from $M$.

b) For $A$ and $B$ to spend the least amount of time on the road, they must arrive at $N$ simultaneously, i.e., they must walk the same distance. Indeed, let $A$ walk a distance $x$ and cycle $15-x$, and let $B$ walk $y$ and cycle $15-y$. The time spent by $A$ is $\frac{x}{6} + \frac{15-x}{15} = 1 + \frac{x}{10}$, and the time spent by $B$ is $1 + \frac{y}{10}$. We are interested in the larger of these two numbers; it should be the smallest possible. In the same amount of time that $A$ walks the distance $x$, $B$ and $C$ can cycle the distance $(15-y) + (15-y-x)$. Therefore, $\frac{30-x-2y}{15} = \frac{x}{6}$, i.e., $7x + 4y = 60$.

The line $x = y$ intersects the line $7x + 4y = 60$ at the point $\left(\frac{60}{11}, \frac{60}{11}\right)$. For any other point on the line $7x + 4y = 60$, the coordinate $x$ or the coordinate $y$ will be larger.

The problem of when $C$ should leave so that $A$ and $B$ arrive simultaneously is exactly the same as problem a).