Which variable should I solve for first?

Solve for the one the question asks about, or whichever makes substitution easier. Here a + c = 40 gives c = 40 - a immediately, and then only a remains in the revenue equation.

How can I check the answer fast?

Multiply back. 14 adult tickets at 5 dollars is 70 dollars, and 26 child tickets at 3 dollars is 78 dollars. The total is 148 dollars, and 14 + 26 = 40, so both conditions hold.

Digital SAT Math: 40 Tickets Sold for 148 Dollars Total, Adult 5 Dollars and Child 3 Dollars. How Many Adult Tickets?

Exact Question Text

A school play sold 40 tickets in total. Adult tickets cost 5 dollars each and child tickets cost 3 dollars each, and the total revenue from ticket sales was 148 dollars. How many adult tickets were sold?

A) 12 B) 14 C) 26 D) 28

1. Step-by-Step Formal Solution

Define variables: $a$ = number of adult tickets, $c$ = number of child tickets.

Translate each sentence into an equation:

$$ a + c = 40 \qquad\text{(total tickets)} $$

$$ 5a + 3c = 148 \qquad\text{(total revenue in dollars)} $$

Substitute $c = 40 - a$ into the second equation:

$$ 5a + 3(40 - a) = 148 $$

$$ 5a + 120 - 3a = 148 $$

$$ 2a = 28 \implies a = 14 $$

Then $c = 40 - 14 = 26$.

Check: $5(14) + 3(26) = 70 + 78 = 148$ ✓

Answer: B

2. Elimination Version

Multiply the first equation by 3: $3a + 3c = 120$. Subtract it from $5a + 3c = 148$:

$$ 2a = 28 \implies a = 14 $$

Same result, and the $c$ terms cancel without any distributing.

3. Fast Desmos Shortcut

Type both equations as a + c = 40 and 5a + 3c = 148 using $x$ and $y$ (x + y = 40, 5x + 3y = 148). The two lines meet at one point, $(14, 26)$. Make sure you read the coordinate that matches the variable the question asks for.

4. Distractor Trap Analysis

C) 26: the number of child tickets. The system is solved correctly but the wrong variable is reported.
A) 12: an arithmetic slip when distributing, for example $5a + 120 - 3a = 148$ read as $2a = 24$.
D) 28: the value of $2a$, the step before dividing by 2.

5. Sanity Estimate Before Solving

If all 40 tickets were child tickets, revenue would be $40 \times 3 = 120$. The actual revenue is 28 more. Each adult ticket adds $5 - 3 = 2$ over a child ticket, so $28 / 2 = 14$ adult tickets. This estimate lands on the answer in two lines.

6. Two More to Practice (Same Method)

30 tickets sold for 110 dollars total, priced at 4 dollars and 2 dollars. If all 30 were 2-dollar tickets, revenue would be 60 dollars. The extra 50 dollars comes 2 dollars at a time from the pricier tickets, so 25 tickets cost 4 dollars and 5 tickets cost 2 dollars.
25 coins, all nickels (5 cents) and dimes (10 cents), worth 190 cents in total. If all were nickels the value would be 125 cents. The extra 65 cents comes 5 cents at a time, so there are 13 dimes and 12 nickels.

For more linear-system practice, see the system with infinitely many solutions.