Answer:
[tex]\large \boxed{A = \$5.50; C = \$3.50}[/tex]
Step-by-step explanation:
Let A = cost of adult ticket
And C = cost of child's ticket. Then,
[tex]\begin{array}{rcll}(1) \qquad 3A + 2C & = & 23.50&\\(2) \qquad 2A + 4C & = &25.00&\\(3) \qquad 6A + 4C & = &47.00&\text{Doubled (1)}\\4A & = & 22.00&\text{Subtracted (2) from (3)}\\(4) \qquad \qquad \quad A & = & \mathbf{5.50}&\text{Divided each side by 4}\\\end{array}\\[/tex]
[tex]\begin{array}{rcll}11.00 + 4C & = &25.00& \text{Substituted(4) into (2)}\\4C &=& 14.00&\text{Subtracted 11.00 from each side}\\C &=& \mathbf{3.50}&\text{Divided each side by 4}\\\end{array}\\\text{The ticket costs are $\large \boxed{\mathbf{A = \$5.50; C = \$3.50}}$}[/tex]
Check:
[tex]\begin{array}{cccl}3(5.50) + 2(3.50) = 23.50 & \qquad & 2(5.50) + 4(3.50) = 25.00\\16.50 + 7.00 = 23.50 & \qquad & 11.00 + 14.00 = 25.00\\23.50 = 23.50 & \qquad & 25.00 = 25.00\\\end{array}[/tex]
OK.
