Writing and Solving a System from a Word Problem

0:00A school play sold 30 tickets for 296 dollars.

0:05Adults pay 12 dollars and students pay 8 dollars.

0:09How many of each kind were sold?

0:12Step 1: name the unknowns. Let a be the number of adult tickets, and s the number of student tickets.

0:21A common mistake is to use the prices as the unknowns, but the prices are already known.

0:28Step 2: write an equation for each fact. The tickets add up to 30, so a plus s equals 30.

0:37Adults bring in 12 times a dollars, written 12a. Students bring in 8 times s.

0:45Together that is 296 dollars: 12a plus 8 times s equals 296.

0:54Together they form a system of linear equations: each graph is a straight line. The solution makes both equations true.

1:03Step 3: solve. Subtract a from both sides of the first equation: s equals 30 minus a.

1:12Substitute, or replace, s in the second equation with 30 minus a.

1:19Expand the brackets: 8 times 30 is 240, and 8 times minus a is minus 8a.

1:28Collect like terms, the terms with a: 12a minus 8a is 4a.

1:35Subtract 240 from both sides: 296 minus 240 is 56, so 4a equals 56.

1:46Divide both sides by 4: 56 divided by 4 is 14, so a equals 14.

1:54Now put 14 in for a: s equals 30 minus 14, which is 16.

2:01Step 4: check both equations. 14 plus 16 is 30.

2:0812 times 14 is 168, 8 times 16 is 128, and 168 plus 128 is 296.

2:20So the play sold 14 adult tickets and 16 student tickets.

2:25To recap: name the unknowns, write an equation for each fact, solve, then check.