To learn how to set up and solve this sort of "uniform rate" exercise, try

**here**.

Then use the standard set-up, noting that "meeting" here actually means "passing". The rate for X, given that the distance from home to the factory is "d", is d/20 units per minute; the rate for Y is d/30 units per minute. Then:

. . . . .X:

. . . . .rate: d/20

. . . . .time: t

. . . . .distance: dt/20

. . . . .Y:

. . . . .rate: d/30

. . . . .time: t - 6)

. . . . .distance: d(t - 6)/30

Since they covered the same distance (from home to when X, having started later but moving faster, caught up to Y), set the two "distance" expressions equal. Since d does not equal zero (they do not live in the factory), you can divide through.

Solve the resulting linear equation.