Question

14. At the museum, the O'Rourke family

Answer 1

Answer:

$\large \boxed{A = \ 5.50; C = \ 3.50}$

Step-by-step explanation:

Let A = cost of adult ticket

And C =  cost of child's ticket. Then,  

$\begin{array}{rcll}(1) \qquad 3A + 2C & = & 23.50&\\(2) \qquad 2A + 4C & = &25.00&\\(3) \qquad 6A + 4C & = &47.00&\text{Doubled (1)}\\4A & = & 22.00&\text{Subtracted (2) from (3)}\\(4) \qquad \qquad \quad A & = & \mathbf{5.50}&\text{Divided each side by 4}\\\end{array}$

$\begin{array}{rcll}11.00 + 4C & = &25.00& \text{Substituted(4) into (2)}\\4C &=& 14.00&\text{Subtracted 11.00 from each side}\\C &=& \mathbf{3.50}&\text{Divided each side by 4}\\\end{array}\\\text{The ticket costs are \large \boxed{\mathbf{A = \ 5.50; C = \ 3.50}} }$

Check:

$\begin{array}{cccl}3(5.50) + 2(3.50) = 23.50 & \qquad & 2(5.50) + 4(3.50) = 25.00\\16.50 + 7.00 = 23.50 & \qquad & 11.00 + 14.00 = 25.00\\23.50 = 23.50 & \qquad & 25.00 = 25.00\\\end{array}$

OK.