Algebra. Systems of equations. Setting up the problem.?

Question

How do I set this word problem up as a system of equations? Lets say:
Up close tickets = C
In the middle tickets = M
Far out tickets = F

Three kinds of tickets are available for a concert: "up close," "in the middle," and "far out." "Up close" tickets cost $10 more than "in the middle" tickets, while "in the middle" tickets cost $10 more than "far out" tickets. Twice the cost of an "up close" ticket is $20 more than 3 times the cost of a "far out" ticket. Find the price of each kind of ticket.

Answer 1

read the question, and write what it is saying

"Up close" tickets cost $10 more than "in the middle" tickets
c=m+10

"in the middle" tickets cost $10 more than "far out" tickets
m=f+10

Twice the cost of an "up close" ticket is $20 more than 3 times the cost of a "far out" ticket.
2c=20+3f or 2c=3f+20

Twice the cost of an up close
2c

is
=

20 dollars more than
+20

3 times the cost of a "far out"
3f

put them together, solve for and sub out variables
c=m+10
m=f+10
c=(f+10)+10 <---combined equasion by subing 'm'
c=f+20
c-20=f <-----------solved for 'f'

2c=3f+20
2c=3(c-20)+20 <---------subed for 'f'
2c=3c-60+20 <----------multiplyed
2c=3c-40 <----------20-60= -40
2c+40=3c <------flopped it (positives are nice)
40 = c <----------2c off

2(40)=3f+20 <------- sub for 'c'
80=3f+20 <------------multiply
60=3f <---------- -20
20 = f <-----------divide

m=f+10
m=20+10 <----------- sub for 'f'f
m=30

c=40
m=30
f=20

now check
c=m+10
40=30+10
40=40 TRUE

m=f+10
30=20+10
30=30 TRUE

2c=3f+20
2(40)=3(20)+20
80=60+20
80=80 TRUE

I read some other answers, picking numbers out of air doesn't work when problems get more complex. follow a process, everytime.(even if it means more typing/writing) that's what algebra teachers want to see, the process. some don't even care if you add wrong, as long as the process is right.

Answer 2

C = M + 10
M = F + 10
2C = 3F + 20

plut C of the first equation for C in the third one
2(M+10) = 3F + 20

F = M - 10

2(M+10) = 3(M - 10) + 20
2M + 20 = 3M - 30 + 20
20 = M - 10
M = 30


F = 30 - 10
F = 20

C = 30 + 10
C = 40

Answer 3

This is what I come up with:
C= 10 + M
M = 10 + F
2C-3F = 20

2(40) - 3(20) = 20
80 - 60 = 20
C= 40, F = 20
therefore
40 = 10 + M M = 30

I hope this helps.

Answer 4

C=M+10
M=F+10
2C=20+3F
F=M-10

Answer 5

okay, the number lines above get you to the solution, but how do you set up the answer? Be sure to include units on your letter (variables) and numbers (knowns).

C tickets = M tickets + $10
etc

This way when you get an answer you'll know what units are attached to it. You can only add or subtract numbers with like units (dollars from dollars) and if it makes sense instead of just manipulating numbers and letters that could be anything.
Hope this helps--use units!

Answer 6

c=m+10
m-10=f
2c=20+3f

m=30
c=40
f=20

Answer 7

C=M+10
C=F+20
M=F+10

2C+20=3F
2(F+20)+20=3F
2F+40+20=3F
F=60
C=80
M=70

kinda did it in my head so definitely check my answer!