Travis has a total of $60 in dimes and quarters. if he could switch the numbers of dimes with the number of quarters, he would have $87. how many of each coin dose he have?

For this, L and T are going to be the two amounts that the dimes and quarters had. We're are going to use systems (two equations) to solve this and we're are going to use elimination.
The equations for this would be:
.1T+.25L=60
.25T+.1L=87
To eliminate, let's multiply the second equation by -4 and the first by 10 so we can eliminate T. 
1T+2.5L=600
-1T-.4L= -348
2.1L=252
252÷2.1= 120 = L
One of the amounts is 120, let's plug it in to find the other amount.
.25T+.1(120)= 87
.25T+12=87
.25T=75
75÷.25=300=T
Let's see if we're right by plugging it in.
300(.25)+ 120(.1)
75+12
87
300(.1)+ 120(.25)
30+30
60
He has 300 dimes and 120 quarters.

The last one bc that's a calendar