Andrew M. answered 07/15/15
Tutor
New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors
We can set up two equations in two variables and solve the system.
Let A = # adults admitted
Let C = # children admitted
1) Since there were 302 people total we know A+C = 302
2) Children cost $1.50 and adults cost $4 and the total fees were so $828 so
4A + 1.5C = 828
Our equations are:
A + C = 302 eqn 1
4A + 1.5C = 828 eqn 2
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We can solve either by substitution or elimination. I will use substitution
Solve for A in terms of C from eqn 1 and substitute that into eqn 2 to solve for C
A = 302-C from eqn 1
4(302-C) + 1.5C = 828
1208 - 4C + 1.5C = 828
-2.5C = 828 - 1208
C = -380/(-2.5) = 152
There were 152 children
The number of adults = 302-C = 302-152 = 150
Check Answer: 4A + 1.5C = 828
4(150) + 1.5(152) = 828
600 + 228 = 828
828 = 828
The answer checks as valid