Question

A ticket to a concert costs either $12 or $15. A total of 300 tickets are sold, and the total money collected is $4140 . The number of $12 tickets sold are?​

Answer 1

Answer:

Step-by-step explanation:

system of equations… so, in the first one, let x =amount of ticket A, let y = amount of ticket B.

So, x + y = 300 tickets

In equation two, we know $12 per ticket A, and $15 per ticket B, but we don’t know the amount of tickets.

So, $12x + $15y = $4,140.

So we can solve this using elimination, substitution, graphing it etc. etc.

I will use substitution, as I prefer that when I have one equation that is x +/- y =

so i prefer multiplying by 15, so I am going to solve for y.

x + y = 300

-x ……. -x

y = -x +300

No substitute the y value into the second equation…

$12x +$15(-x + 300) =$4,140.

so we distribute the 15

12x -15x + 4500 = 4140

combine like terms:

-3x +4500 = 4140

subtract 4500 from both sides

-3x = -360

divide both sides by -3

x = 120. So we know that there were 120 of tickets A bought

so we go back to our system and solve

x + y = 300, but use our newly acquired x value.

120 + y = 300

y = 180

So, 120 of ticket A, 180 of ticket B, and we could plug it in and get the same answer….. just for fun

12x + 15y = 4140

12 (120) + 15 (180)

1440 + 2700 = 4140.

Answer 2

$120$ tickets

If $a,b,r$ are real numbers (and if $a,b$ are not both equal to $0$) then $ax+by = r$ is called a linear equation in two variables.

Number of tickets that cost $\ 12=x$

Number of tickets that cost $\ 15=y$

Total number of tickets $=300$

$x+y=300$

      $y=300-x$

Total cost of all the tickets $=\ 4140$

$12x+15y=4140$

Put $y=300-x$

$12x+15(300-x)=4140$

 $12x+4500-15x=4140$

                         $3x=360$

                           $x=120$

So, $\boldsymbol{120}$ tickets are sold that cost $\ 12$ each.

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