Peter R. answered 02/22/23
Experienced Instructor in Prealgebra, Algebra I and II, SAT/ACT Math.
You set up a system of equations, one for the revenue on day 1 and another for day 2.
Day 1: 6c + 10a = 150 (c = price of child's ticket; a = price of adult's ticket)
Day 2: 15c + 5a = 135
These can be solved by elimination or substitution, but they specify graphing.
As they are straight lines, the easiest way is to find two points on the graph for each equation and draw a line between them.
Easiest way to do that is to set c = 0 and solve for a, then set a = 0 and solve for c for each equation.
Ex: Day 1: 6(0) + 10a = 150, so a = 15 (point 0,15); 6c + 10(0) = 150, so c = 25 (point 25,0).
When both lines are drawn, intersection is the solution. Check with Desmos Graphing Calculator website.