Hilton university drama club sold 311 tickets for a play. student tickets cost 50 cents each, nonstudent tickets cost $1.50. if total receipts were 385.50, find how many tickets of each type were sold

X= number of student tickets

Y= number of on student tickets

x+y=311

0.50x + 1.50y = 385.50

Solve x in 1st equation

X+y=311

X= -y + 311

Now substitute x into second equation to solve for y

0.50(-y+311)+ 1.50y = 385.50

-0.50y+311.50+1.50y=385.50

Y+311.50=385.50

Y= 385.50-311.50

Y= 74

Now substitute y answer to first equation to find x

X+y=311

X+ 74=311

X= 311-74

X= 237

So 237 students tickets and 74 non students tickets

step-by-step explanation:

quartic functions has x = –1 and x = –2 as its only two real zeroes:

f(x) = a(x+2)(x+1)   a: real number

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!