Lets set the problem up.
Let s represent the senior tickets
Let c represent the children tickets
Translating into algebra: She sold 4 senior tickets is 4s, 5 senior tickets 5s
She sold 2 children's ticket 2c, 7 children's tickets 7c
We are almost there, Jeff. But next, we need to set up the system of linear equations from the information provided.
Now that we have the equations, we need to decide which method is the easiest to solve for x and y. In this case, in my opinion, the elimination method is the easiest. And probably getting rid of the s is the best way to approach this problem. Again, this is my opinion. I am going to eliminate the s's by multiplying the first equation by -5 and the second equation by 4. If I do that, I will end up with -20s and 20s. These two will cancel each other out. Ok, lets do it.
-5(4s+2c=80)= -20s-10c= -400
4(5s+7c=145)= 20s+28c= 580
So, children's tickets price is 10. Next, I am going to plug in 10 for c into the first original equation. Why?? The numbers are smaller.. :).
The price for the senior citizen ticket is 15.
So, it is 10 for the children's tickets and 15 for the senior's tickets..
Ok, Jeffrey.. Glad to help.. and don't fall behind.. And please rate my answer.
D. Y. Taylor