SOLUTION: The cost of admission to the Christmas Carnival was $50 for a group of 9 children and 2 adults. The admission was $60 for another group of 10 children and 4 adults. What was the a

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: The cost of admission to the Christmas Carnival was $50 for a group of 9 children and 2 adults. The admission was $60 for another group of 10 children and 4 adults. What was the a      Log On


   

Question 1188683: The cost of admission to the Christmas Carnival was $50 for a group of 9 children and 2 adults. The admission was $60 for another group of 10 children and 4 adults. What was the admission price for each child? What was the admission price for each adult?
Found 2 solutions by Shin123, Boreal:
Answer by Shin123(626) About Me  (Show Source):
You can put this solution on YOUR website!
Letting x be the cost of admission for 1 child, and y for 1 adult, we have system%289x%2B2y=50%2C10x%2B4y=60%29
Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
+system%28+%0D%0A++++9%5Cx+%2B+2%5Cy+=+50%2C%0D%0A++++10%5Cx+%2B+4%5Cy+=+60+%29%0D%0A++We'll use substitution. After moving 2*y to the right, we get:
9%2Ax+=+50+-+2%2Ay, or x+=+50%2F9+-+2%2Ay%2F9. Substitute that
into another equation:
10%2A%2850%2F9+-+2%2Ay%2F9%29+%2B+4%5Cy+=+60 and simplify: So, we know that y=2.5. Since x+=+50%2F9+-+2%2Ay%2F9, x=5.

Answer: system%28+x=5%2C+y=2.5+%29.

Therefore, the admission for a child cost $5, and an adult $2.50.

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
9C+2A=50; C=children, A=adult
10C+4A=60
multiply the top by -2 and add
-18C-4A=-100
10C+4A=60
-8C=-40
C=$5 for children
substitute into first
45+2A=-0
A=$2.50
-
check using second
10(5)+4(2.50)=60, and that is true.