Question

Brenda's school is selling tickets to a spring musical. On the first day of ticket sales the school sold 3
senior citizen tickets and 9 child tickets for a total of $75. The school took in $67 on the second day by
selling 8 senior citizen tickets and 5 child tickets. What is the price each of one senior citizen ticket and
one child ticket?
@iambatman @iGreen @e.mccormick

Answer 1

OK, so, how do you think you need to set it up? Obviously you need to make two equations to find the two unknowns.

Answer 2

Right thats what i dont know how to do I can't solve these without a verbal model

Answer 3

So, lets take it one line at a time.

On the first day of ticket sales the school sold 3 senior citizen tickets and 9 child tickets for a total of $75.

So what numbers and what variables do you see there?

Answer 4

3 senior tickets
9 child tickets
total $75

Answer 5

Good. So, if we just shorten the names we can say:

3s
9c
total $75

Now, can you put that into an equation?

Answer 6

3s+9c=75?

Answer 7

Yep! Exactly! Now, do the same basic thing for " The school took in $67 on the second day by selling 8 senior citizen tickets and 5 child tickets."

Answer 8

8c+5c=67?

Answer 9

oops

Answer 10

wait no 8s+5c=67?

Answer 11

Not both c.

Answer 12

OK, now you have:
3s+9c=75
8s+5c=67

Answer 13

so now all i do is solve right?

Answer 14

Yep!

Answer 15

ok i have it from here thanks! :D

Answer 16

senior citizen ticket: $4, child ticket: $7 is this right?

Answer 17

Yah.

Answer 18

I'll show some other ways to solve it for the fun of it...

Answer 19

ok :)

Answer 20

3s+9c=75
8s+5c=67

Lets see... I see three easy solutions to that. First, graphing:

https://www.desmos.com/calculator/laerywuegn

Now for some elimination in a matrix!

\(\left[\begin{array}{cc|c}
3 & 9 & 75\\
8 & 5 & 67\\
\end{array}\right]\)

divide row one by 3.

\(\left[\begin{array}{cc|c}
1 & 3 & 25\\
8 & 5 & 67\\
\end{array}\right]\)

Multiply row 1 by -8 and add to row 2.

\(\left[\begin{array}{cc|c}
1 & 3 & 25\\
0 & -19 & -133\\
\end{array}\right]\)

Divide row 2 by -19.

\(\left[\begin{array}{cc|c}
1 & 3 & 25\\
0 & 1 & 7\\
\end{array}\right]\)

Multiply row 2 by -3 and add to row 1:

\(\left[\begin{array}{cc|c}
1 & 0 & 4\\
0 & 1 & 7\\
\end{array}\right]\)

Presto!

And then there is basic substitution

3s+9c=75
8s+5c=67

3s=75-9c
s=25-3c

So:

8(25-3c)+5c=67
200-24c+5c=67
-19c=67-200
-19c=-133
c=7

3s+9(7)=75
3s+63=75
3s=12
s=4

Answer 21

Took me more time to type up the augmented matrix than to do the math. LOL

Answer 22

O_O wow theres so many ways to do one problem O_O i used elimination but i think the substituition way its alot more easier :) you make it look like a picece of cake xD

Answer 23

Well, when I saw that all the parts of the first equation were dividible by 3...

Answer 24

Ah ok that makes sense that would've been easier to do XD thanks :D