Mofor’s school is selling tickets to the annual dance competition. on the first day of ticket sales the school sold 7 adult tickets and 6 tickets for a total of $143. the school took in $187 on the second day by selling 4 adult tickets and 13 student tickets. find the price of an adult ticket and the price of a student ticket.

The price of one adult ticket is $ 11 and the price of one student ticket is $ 11

Given that , Mofor’s school is selling tickets to the annual dance competition.  

Let the cost of one adult ticket be $m and the cost of one student tickets be $n.

On the first day of ticket sales the school sold 7 adult tickets and 6 tickets for a total of $143.  

7m + 6n = 143 ------- eqn (1)

The school took in $187 on the second day by selling 4 adult tickets and 13 student tickets.  

4m + 13n = 187  ------ eqn (2)

We have to find the price of an adult ticket and the price of a student ticket.

Now, let us solve the equations.

Multiply eqn 1 by 4

28m + 24n = 572  ----- eqn 3

Multiply eqn 2 by 7

28m + 91n = 1309  ---- eqn 4

Now subtract eqn 4 from eqn 3

28m + 24n = 572

28m + 91n = 1309

(- )--------------------------------------

– 67n = - 737

67n = 737

n = 11

Plug in n = 11 in eqn 1

7m + 66 = 143

7m = 143 – 66

m = 11

Hence, the cost of one adult ticket is $ 11 and cost of one student ticket is $11

step-by-step explanation:

just plug in each number for x and see which ones work and don't

aloha~! my name is zalgo and i am here to you out on this amazing day. to answer your question, yes, nonzero numbers can have a greatest common factor and that would be 1. still don't get it? let me considering every number is divisible by 1 and hence every number has 1 as a factor of theirs. so, if 2 nonzero numbers didn't share any gcf (greatest common factor), they would still have a factor in common and that would be 1.

i hope that this ! : p

"stay and stay proud! " - zalgo

(by the way, do you mind marking me as brainliest? i'd greatly appreciate it! mahalo! x3)