The enrollment in college a may be modeled by y = 0.046x + 0.570, and the enrollment in college b may be modeled by y = –0.036x + 2.702, where x is the number of years since 1990 and y is the enrollment in thousands. when will the two colleges have the same enrollment and what is that enrollment?     in 1974, both colleges had an enrollment of 1,766 students.     in 2016, both colleges will have an enrollment of 1,766 students.     in 2012, both colleges will have an enrollment of 1,122 students.     in 1992, both colleges had an enrollment of 1,600 students.

You want to know when both colleges have the same enrollment. "When" is a time thing so you are going to be solving for x. The number of students is the same and that means you will be solving for y.

Since both ys are equal, you can equate the right side of each equation to each other.

0.046 x + 0.570 = - 0.036x + 2.702 There are a number of ways to go on. The easiest is to dig out your calculator. Add 0.036x to both sides.
0.046x + 0.036x + 0.570 = 2.702
0.082x + 0.570 = 2.702    Now subtract 0.570 from both sides.
0.082x = 2.702 - 0.570
0.082x = 2.132 Divide by 0.084
x = 2.132 / 0.082
x = 26 which means you add 26 onto 1990. The year this took place was 2016 
x = 2016 (That's the year there was equality in enrollment). The second one is the only year that gives 2016 as an answer. So you don't have to find y. But we'll do it anyway. 

Now you have to solve for y
y = 0.046x +0.57 put 26 in for x
y = 0.046 * 26 + 0.570
y = 1.196 + 0.570
y = 1.766 enrollment numbers were equal, but this is in thousands.
y = 1766 enrollment in actual numbers of students.

Second choice <<<<< answer.