Step-by-step explanation:
Let a be the price of 1 adult ticket.
Let c be the price of 1 child ticket.
given,
[tex]6a + 2c = 124[/tex]
as equation 1,
and
[tex]3a + 5c = 100[/tex]
as equation 2.
Now we will solve for a and c using elimination method of simultaneous equations.
Now we multiply equation 2 by 2 to eliminate a and solve for c.
[tex]3a \times 2 + 5c \times 2 = 100 \times 2 \\ 6a + 10c = 200[/tex]
This new equation will be equation 3.
Now we will use equation 1 - equation 3 to eliminate a and solve for c.
[tex](6a - 6a) + (2c - 10c) = 124 - 200 \\ 0 + ( - 8c) = - 76 \\ - 8c = - 76 \\ c = - 76 \div - 8 \\ = 9.5[/tex]
Now substitute c into equation 2.
[tex]3a + 5(9.5) = 100 \\ 3a + 47.5 = 100 \\ 3a = 100 - 47.5 \\ 3a = 52.5 \\ a = 52.5 \div 3 \\ = 17.5[/tex]
Therefore one adult ticket will cost $17.50 and one child ticket will cost $9.50.