Answer:
Your answers should be a = 135 adult tickets and s = 186 student tickets.
Explanation:
1. Solving for variable s by manipulating variable a:
a + s = 321
a = 321 - s (subtract s to the other side to get a by itself)
$3.50(321-s) + $2.50s = $937.50 (substitution from step 2)
$1123.50 - $3.50s + $2.50s = $937.50 (distributive property)
$1123.50 - $1.50s = $937.50 (combine like terms)
-$1.00s = -$186 (subtraction property of equality)
s = 186 student tickets.
2. Solving for variable a by manipulating variable s:
a + s = 321
s = 321 - a
$3.50a + $2.50(321-a) = $937.50
$3.50a + $802.50 - $2.50 a = $937.50
$1.00a = $135
a = 135 adult tickets.
If you plug both numbers into the original equations, they should come out to be true equations. 135 + 186 = 321 and $472.5 + $465 = $937.5, so these two numbers are your answers for a and s accordingly.
Using the elimination method, we need to eliminate one of the variables by a process of subtraction or addition depending on the case. We can use multiplication or division before eliminating in order to eliminate.
a + s = 321
3.50a + 3.50s = 1123.50 (multiply entire equation by 3.50 in order to eliminate variable a by subtraction)
Now, we can use elimination:
3.50a + 3.50s = 1123.50
3.50a + 2.50s = 937.50
0a + 1s = 186, or s = 186 student tickets.
The same process can be used to solve for variable a. Your answers should be a = 135 adult tickets and s = 186 student tickets.