Adult tickets for a play cost 25$ each and childrens tickets cost 15$ each. The total receipts were $7,200 and the total attendance was 400. How many adults and how many children attended? plzz help!
set up the two equations. do you know what they are
hmm not really
two unknowns. call them x and y they represent?
the adults and the children right?
yes and we know how many total attended. that would be the sum of the two. we also know how much each ticket cost and how much total money was brought in. call adults x and children y. can you get the two equations yet?
hmm x+25=400? and y+15=400?
x+y=400, the number of adults+children=400 25x+15y=7200, the adult tickets cost $25 and the children cost $15, and the total moey brought in by all tickets is $7200. making sense?
*money
o yea now it does now wat do we do?
eliminate one of the variables. to do this you can add/subtract multiples of equations or solve for one variable in terms of the next. 2 equations, 2 unknowns.
ok im getting it alittle bit of it
use the simplest equation to find x in terms of y and a constant. plug that value of x into the second equation, for x. this gives you an equation with only y and constants. solve for y, plug that y value into the equation for x.
ok so wat would it be?
so i would plug 400 on the equation 25x+15y=7200?
x=400-y 25(400-y)+15y=7200 solve this for y, then plug that value of y into: x=400-y
ok let me do it and c wat i get
when you are done, x and y should add up to 400 and 25x+15y should be 7200, if we did it right
for y i get 13 am i right?
25*400-25y+15y=7200 -10y=7200-25*400
so it would be 10-y=-2800? am i right
-10
-10y, but yes. now y=?
280 positive right?
correct, the negatives on both sides cancel. x=400-y
it would be 120 right?
then verify that 25(120)+15(280)=7200.
yes it does =)
Nice job
thank you!!!
your welcome.
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