A total of 54 adults and students show up for the play. Adult tickets cost $8 and kid tickets cost $5 and the total money collected was $360. How many of each type of ticket were sold?
It’s using substitution or elimination to find the number of kids and adults. Really need help!!

9514 1404 393

Step-by-step explanation:

If find that "substitution" works well for "mixture" problems like this. Usually, you will want to substitute for the lower-priced item, so that you're solving for the number of higher-priced items.

Let "a" and "k" stand for the number of adult and kid tickets, respectively.

  a + k = 54 . . . . . . total number of tickets sold

  8a +5k = 360 . . . total revenue

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To substitute for k, we solve the first equation for k:

  k = 54 -a

Now, we substitute that into the second equation:

  8a +5(54 -a) = 360

  3a = 90 . . . . . . . . . . . subtract 270 from both sides; collect terms

  a = 30 . . . . . . . . . . . . divide by 3

  k = 54 -30 = 24 . . . . . use the above formula for k

The number of kids was 24; the number of adults was 30.

idk sry

step-by-step explanation:

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions