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one night a theater sold 548 movie tickets. An adults ticket costs 6.50 and a child's ticket cost 3.50. in all 2881 was taken in. how many of each kind of ticket were sold

User MrMesees
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1 Answer

3 votes

Answer:


\large \boxed{\text{321 adult tickets and 227 child tickets}}

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:


\begin{cases}(1) & a + c = 548\\(2) & 9.70a + 5.90c = 2881\end{cases}

2. Solve the equations


\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 & = & (963)/(3.00) &\text{Divided each side by 3.00}\end{array}


\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 $\large \boxed{\textbf{321 adult tickets and 227 child tickets}}$}

3. Check


\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}

OK.

User Yasitha
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