adult tickets at a football game cost $6 and student tickets cost $2. A total of $1776 was collected on 372 tickets. How many ad
2 answers:
258 adult tickets were sold.
1) Set up your 2 equations
• 6a +2s= $1776 (remember cost of each type of ticket goes with the total cost)
• a+s= 372 (number of tickets goes with the total number of tickets)
2) Easy way is to use your calculator and the matrix option
(first section: two rows/two columns)
(second section: two rows/one column)
•put the first section to the negative 1 power and multiply it by the second section
-> The answer for # of adult tickets are on top and student is on bottom. I have an example on the bottom
Or a second way
1) Set up equations
• 6a+2s=$1776
• s+a= 372
2) Rearrange bottom equation to use substitution method
• a= 372- s
3) Then substitute a with top equation and distribute
• 6(372-s) + 2s= $1776 = 2232-6s + 2s= $1776
4) Add like terms and solve for s
• 2232-4s= $1776
-> minus 2232 from 1176= -456
-> divide -4s by itself and -456
-> number of student tickets are 114
5) Plug in 114 to equation (a=372-s) Then subtract 372- 114 and you get the number of adult tickets! :)
Hopefully this helped <3
1) Set up your 2 equations
• 6a +2s= $1776 (remember cost of each type of ticket goes with the total cost)
• a+s= 372 (number of tickets goes with the total number of tickets)
2) Easy way is to use your calculator and the matrix option
(first section: two rows/two columns)
(second section: two rows/one column)
•put the first section to the negative 1 power and multiply it by the second section
-> The answer for # of adult tickets are on top and student is on bottom. I have an example on the bottom
Or a second way
1) Set up equations
• 6a+2s=$1776
• s+a= 372
2) Rearrange bottom equation to use substitution method
• a= 372- s
3) Then substitute a with top equation and distribute
• 6(372-s) + 2s= $1776 = 2232-6s + 2s= $1776
4) Add like terms and solve for s
• 2232-4s= $1776
-> minus 2232 from 1176= -456
-> divide -4s by itself and -456
-> number of student tickets are 114
5) Plug in 114 to equation (a=372-s) Then subtract 372- 114 and you get the number of adult tickets! :)
Hopefully this helped <3
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one way would be to factor
I can't factor it so we will have to use Descartes' Rule of Signs which is helpful for finding how many real roots you have
it goes like this:
for a polynomial with real coefients, consider .
after arranging the terms in decending order in terms of degree, count how many times the signs of the coeffients change direction and minus 2 from that number until you get to 1 or 0. that will be the number of even roots the function can have
We have (-, -, +, +). the signs changed 1 times, so it has 1 real positive root
to get the negative roots, we evaluate f(-x) and see how many times the root changes
signs are (-, -, -, +). there was 1 change in sign
so the function has 1 real negative root
a total of 2 real roots
a function of degree can have at most,
roots
our function is degree 6 so it has 6 roots
if 2 are real, then the others must be complex
6-2=4 so there are 4 complex roots
you can also show that there are only 2 real roots by using a graphing utility to see that there are only 2 real roots