Answer:
[tex]\large \boxed{\text{321 adult tickets and 227 child tickets}}[/tex]
Step-by-step explanation:
1. Set up the equations
Let a = the number of adult tickets
and c = the number of child tickets. Then
6.50a = revenue from adult tickets and
3.50c = revenue from child tickets
6.50a + 3.50c = total ticket revenue
You have a system of two equations:
[tex]\begin{cases}(1) & a + c = 548\\(2) & 9.70a + 5.90c = 2881\end{cases}[/tex]
2. Solve the equations
[tex]\begin{array}{lrcll}(3) & c& = &548 - a&\text{Subtracted a from each side of (1)}\\ & 6.50a + 3.50(548 - a) & = & 2881 &\text{Substituted (3) into (2)}\\& 6.50a + 1918 - 3.50a & = & 2881 &\text{Distributed the 3.50}\\&3.00a + 1918 & = & 2881 &\text{Combined like terms}\\& 3.00a & = & 963 &\text{Simplified}\\& a & = & \dfrac{963}{3.00} &\text{Divided each side by 3.00}\end{array}[/tex]
[tex]\begin{array}{lrcll}(4) &a & = & \mathbf{321} &\text{Simplified}\\ &321 + c & = & 548 &\text{Substituted (4) into (1)}\\ &c & = & \mathbf{227} &\text{Subtracted 321 from each side}\\\end{array}\\\text{The museum sold $\large \boxed{\textbf{321 adult tickets and 227 child tickets}}$}[/tex]
3. Check
[tex]\begin{array}{rclcrcl}321 + 227& = & 548 & \qquad &6.50(321) + 3.50(227)&=&2881\\548 & = & 548 & \qquad &2086.50 + 794.50& = &2881\\&&& \qquad &2881& = &2881\\\end{array}[/tex]
OK.
