There's actually a second way to do the problem that introduces systems of equations:

First, you write an equation for the number of seats. Say that "x" is the number of seats that sit 2 children and "y" is the number of seats that sit 3 children. You know that x + y = 7.

Furthermore, you know that 2 students sit in each "x" seat and 3 in each "y" seat, and the total must be 19 children, so you can say 2x + 3y = 19.

Now you have two equations: x + y = 7 and 2x + 3y = 19.

Isolate one variable, let's say x, from x + y = 7 by subtracting y from both sides and getting x = 7-y. Now substitute that into the second equation.

So 2x + 3y = 19, and x = 7 - y. So you have 2(7-y) + 3y = 19.

Multiply that out to get 14 - 2 y + 3y = 21. Simplify by adding like terms to get 14 + y = 19. Then subtract 14 from both sides, so y = 5.

Then plug the y =5 into the first equation, x + y = 7. You have x + 5 = 7. Subtract 5 from both sides to get x = 2.

Then you know that you have 5 rows that are "y's" and seat 3 children, and 2 rows that are "x's" and seat 2 children.

Hope that helps as another way to think about the problem!

if you seat 2 children to one seat for all 7 seats you have 2*7=14 children which leaves 19-14=5

children without a seat

therefore each of the 5 children will sit in a seat with 2 other children

this means that there are 5 seats with 3 children in them and 2 seats with 2 children in them

5*3=15 and 2*2=4, 15+4=19

or, 3*7=21(3 children to a seat), but there are only 19 children so 2 seats will have only 2 children in them and the other 5 seats will have 3 children in them