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Sagot :
a=adult
c=child
Equation 1 (the first family)
2a+3c=56.5
Equation 2 (the second family)
a+4c=49.5 multiply by 2
2a+8c=99 multiply by -1
-2a-8c=-99
Systems of Equations
(2a+3c=56.5)
+ (-2a-8c=-99)
-------------
-5c=-42.5
c=$8.5
Substitute
2a+3*8.5=56.5
2a+25.5=56.5
2a=31
a=$15.5
Conclusion
Adult Ticket: $15.5
Child Ticket: $8.5
c=child
Equation 1 (the first family)
2a+3c=56.5
Equation 2 (the second family)
a+4c=49.5 multiply by 2
2a+8c=99 multiply by -1
-2a-8c=-99
Systems of Equations
(2a+3c=56.5)
+ (-2a-8c=-99)
-------------
-5c=-42.5
c=$8.5
Substitute
2a+3*8.5=56.5
2a+25.5=56.5
2a=31
a=$15.5
Conclusion
Adult Ticket: $15.5
Child Ticket: $8.5
The cost of an adult ticket is $15.50 and the cost of a child ticket is $8.50
Proof:
Set up a system of equations calling Adults a and Children c
2a+3c= $56.50 (equation 1)
4c+a= $49.50 (equation 2)
a= -4c+49.50 (i subtracted 4c to both sides) (label equation 3)
sub equation 3 into equation 1
2(-4c+49.50)+3c= 56.50
-8c+99+3c=56.50
-5c+99=56.50
-99 -99
-5c = -42.50
c= -42.50/ -5
c= 8.50
sub c=8.50 into equation 3 to find adult price
a= -4(8.5)+49.50
a= 15.50
Therefore Child Cost = $8.50 and Adult Cost= $15.50
Proof:
Set up a system of equations calling Adults a and Children c
2a+3c= $56.50 (equation 1)
4c+a= $49.50 (equation 2)
a= -4c+49.50 (i subtracted 4c to both sides) (label equation 3)
sub equation 3 into equation 1
2(-4c+49.50)+3c= 56.50
-8c+99+3c=56.50
-5c+99=56.50
-99 -99
-5c = -42.50
c= -42.50/ -5
c= 8.50
sub c=8.50 into equation 3 to find adult price
a= -4(8.5)+49.50
a= 15.50
Therefore Child Cost = $8.50 and Adult Cost= $15.50
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