The slow cat puts 10 mice per minute in his basket.

......4 minutes later......

The fast cat wakes up, and puts 20 mice per minute in his basket.

(The slow cat: blue line)

(The fast cat: pink line)

y=10*x in minutes

y=20*(x-4) in minutes

The 10*x and 20*x in these equations are also called: "the slope of the line".

In this grid, each vertical space represents 10 mice; and each horizontal space, represents 1 minute (60 seconds)

SOLUTIONS:

After 8 minutes total, (4 minutes for the fast cat, since he woke after 4 minutes):

The slow cat has 80 mice in his basket (8 min*10 mice/minute)

The fast cat has 80 mice in his basket (4 min*20 mice/minute)

If you want to solve this problem without graphing, here is one way:
The Slow Cat catches mice at a rate of 10 mice/minute; the Fast Cat at a rate of 20 mice/minute.
We're looking for the time at which they have the same number of mice in the 2 baskets.
So we must set the number of mice in both baskets as equal:
10*x=20*(x-4)
10*x=20*x-80
10*x+80=20*x-80+80
10*x+80=20*x
10*x-10*x+80=20*x-10*x
80=10*x
x=80/10
x=8 minutes

The same in seconds:
10 mice/minute = 10/60 = 0.1666666 mice/second
20 mice/minute = 20/60 = 0.3333333 mice/second
4 minutes=240 seconds
0.1666666*x=0.3333333*(x-240)
0.1666666*x=0.3333333*x-80
80=0.1666666*x
x=80/0.1666666
x=480 seconds = 8 minutes