Let x =# of children, y = # of students and z = # of adults.


Then you have this two equation


x + y + z = 1050    (1)   (the total)   and

z =            (2)   ("there were half as many adults as children and students combined.")


Substitute eq(2) into eq(1), replacing z.  You will get


x + y +  = 1050,   or


 = 1050,   which gives you  


x + y =  = 700.


So, you know now that the combined number of children and students was 700.


You also know that the number of adults was  1050 - 700 = 350.


So, adults payed  350*7 = 2450 dollars;

then children and students payed 5150 - 2450 = 2700 dollars for their tickets.



Now you have this system of 2 equations in 2 unknowns


 x +  y =  700,      (2)
3x + 5y = 2700.      (3)


Thus you reduced the original problem from 3 unknowns to only 2 unknowns


Now solve the system (2),(3) by substitution method. From eq(1) express y = 700 - x and substitute it into eq(2). You will get


3x + 5*(700-x) = 2700

3x - 5x = 2700 - 3500

-2x = -800  ====>  x = 400.


The number of children was 400;  the number of students was  700-400 = 300;  the number of adults was  350.