Assume that the first car used x gallons, and the second car used y gallons this week
Since the total number of gallons is 60, then

Since the rate of the first car is 40 miles per gallon
Since the rate of the second car is 20 miles per gallon
Then the first car made (40)(x) miles, and the second car made (20)(y) miles
Since the total distance they made is 1900 miles, then

Now, we have a system of equations to solve it.
Multiply equation (1) by -20 to make the coefficients of y equal and different in signs to eliminate it

Add equations (2) and (3)

Divide both sides by 20

Substitute the value of x in equation (1) to find y

Subtract 35 from both sides

The first car used 35 gallons
The second car used 25 gallons