82 (1192). The distance between points $A$ and $B$ is 60 km. A car leaves from $A$ to $B$, and at the same time, a second car leaves from $B$ in the same direction. If the speed of the first car is increased by 10 km/h, and the speed of the second car is increased by 8 km/h, the first car will catch up with the second at the same place but one hour earlier. What is the speed of each car?
Solution. Method I. Let $t$ hours be the time it takes for one car to catch up with the other when they are driving at their usual speed. Then their rate of approach is $\frac{60}{t}$ kilometers per hour. If the speed of one car increases by 10 km/h and the other by 8 km/h, their rate of approach will increase by 2 km/h $(10-8=2)$ and will be $\frac{60}{t-1}$ kilometers per hour.

We have the equation

$$
\frac{60}{t-1}-\frac{60}{t}=2, \text { from which } t=6
$$

If the speed of the first car is $v_{1}$ kilometers per hour, then in 6 hours it will travel a distance of $6 v_{1}$ kilometers. The same distance with an increased speed of 10 km/h, this car will take 1 hour less, i.e., we have the equation

$$
6 v_{1}=5\left(v_{1}+10\right), \text { from which } v_{1}=50
$$

Similarly, if $v_{2}$ is the speed of the second car, then we have the equation $6 v_{2}=5\left(v_{2}+8\right)$, from which $v_{2}=40$.

Method II. Let the difference in the speeds of the cars be $v$ kilometers per hour. If the speeds of the cars are increased by 10 km/h and 8 km/h, respectively, the difference in speeds will be $(v+2)$ kilometers per hour. Since with the increased speeds, one car will catch up with the other 1 hour earlier, we have the equation

$$
\frac{60}{v}-\frac{60}{v+2}=1, \text { from which } v=10
$$

Since the difference in the speeds of the cars is 10 km/h, one car will catch up with the other in 6 hours $(60: 10=6)$, and with the increased speeds, in 5 hours $(60:(10+2)=5)$. Since the first car will catch up with the second in the same place, we have the equation $6 v_{1}=5\left(v_{1}+10\right)$, where $v_{1}$ is the speed of the first car. From this, $v_{1}=50.50-10=40$ km/h is the speed of the second car.

Answer: 50 km/h; 40 km/h.

$\square 83(1193)$. Construct the graph of the equation:
a) $|x|+|y|=4$; b) $|x|-|y|=4$.

Solution. a) Method I. Based on the definition of the absolute value of a number, we have:

198

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-202.jpg?height=478&width=472&top_left_y=255&top_left_x=358)

Fig. 33

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-202.jpg?height=471&width=349&top_left_y=256&top_left_x=862)

Fig. 34

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-202.jpg?height=474&width=468&top_left_y=251&top_left_x=1225)

Fig. 35

1) For $x \geqslant 0, y \geqslant 0$, we have the equation $x+y=4$.
2) For $x \geqslant 0, y<0$, we have the equation $x-y=4$.
3) For $x<0, y \geqslant 0$, we have the equation $-x+y=4$.
4) For $x<0, y<0$, we have the equation $-x-y=4$.

We construct the graphs of the four obtained equations in the corresponding quadrants (Fig. 33).

Method II. Move $|x|$ to the right side of the given equation, we get $|y|=4-|x|$. We will construct the graph of this equation in the following sequence:

1) $y=4-x$ (Fig. 34)
2) $y=4-|x|$ (Fig. 35)
3) $|y|=4-|x|$ (Fig. 33).

Note. It is advisable to explain to students in advance the rules for constructing graphs of functions $y=f(|x|)$ and graphs of dependencies $|y|=f(x)$.

b) Method I. Based on the definition of the absolute value of a number, we have four equations:

1) $x-y=4$ for $x \geqslant 0, y \geqslant 0$;
2) $x+y=4$ for $x \geqslant 0, y<0$
3) $-x-y=4$ for $x<0, y \geqslant 0$;
4) $-x+y=4$ for $x<0, y<0$.

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-202.jpg?height=405&width=489&top_left_y=1942&top_left_x=361)

Fig. 36

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-202.jpg?height=399&width=335&top_left_y=1945&top_left_x=929)

Fig. 37

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-202.jpg?height=399&width=388&top_left_y=1945&top_left_x=1322)

Fig. 38

By constructing the graph of each of the four equations for the corresponding quadrants, we obtain the graph of the given equation (Fig. 36).

Method II. The graph of the equation $|y|=|x|-4$ will be constructed in the following sequence:

$\begin{array}{ll}\text { 1) } y=x-4 & \text { (Fig. } 37) \text {; } \\ \text { 2) }\end{array}$
2) $y=|x|-4$ (Fig. 38);
3) $|y|=|x|-4$ (Fig. 36).