Your solving a linear equation for this problem the easiest method of the 3 is going to be solving by elimination: 1. The adult tickets (A) coast $13 per adult and student tickets (S) coast $8. The theater sold $568 worth of tickets. because we don't know how many of each tickets were sold we multiply the prices of adults by one variable and the price of students by another variable to add up to a total of $586 hence the equation 13a+8s=568
For the second equation you simply want to add the number of adult tickets and the number of student tickets together to get the total amount of tickets which in this case Is 56 hence the equation a+s=56 13A+8S=568 A+S=56
2. the next step would be to add the equation to solve for either A or Sby first seeing if any of the variables cancel each other out. In this case they don't so you need to multiply one of the equations to make it so one of the variables cancel out. I multiplied the second equation by -13 (I chose the second equation because it's easiest) so you and up with -13(a+b=56) and then using distributive property you and up with the equation -13a-13s=-728 so your new equation ends up as: 13A+8S=568 -13A-13s=-728
3. now you can add the equations together 13a+-13a=0 (cancel each other out) 8s+-13s=-5s and 568+-728=-160 so you and up with -5s=-160
13a+8s=568 +-13a-13s=-728 -5s=-160
4. now you want to isolate s by dividing -5 into -160 which is 32 hence s=32
s=-160÷-5=32 s=32
5. now that you have your value for s you can plug it in to either equation (I used the second because it's easiest) and then to find the value of a you have to isolate a so you subtract 32 from both sides so you get a=56-32 therefore a= 24 a+32=56 a=56-32 a=24
6.You now have both values of your variables the theater sold 32 student tickets and 24 adult tickets
