SOLUTION: I know how to do the problem I just cant figure out how to set the problem up(translate). Thank you. 21.How many gallons of a 12% solution must be mixed with a 20% solution to g

Question 203002This question is from textbook Beginning Algebra
: I know how to do the problem I just cant figure out how to set the problem up(translate). Thank you.
21.How many gallons of a 12% solution must be mixed with a 20% solution to get 10 gallons of a 14% solution.

32.A movie theater has two ticket prices: $8 for adults and $5 for children. If the box office took in $4116 from sales of 600 tickets, how many tickets of each kind were sold? This question is from textbook Beginning Algebra

Found 2 solutions by rfer, scott8148:
Answer by rfer(16322) (Show Source): You can put this solution on YOUR website!
A)
let x=12% solution
let 10-x=20% solution
.12x+.20(10-x)=.14(10)
.12x-.20x=1.4-2
-.08x=-.6
x=7.5
7.5 gal of 12%
2.5 gal of 20%
--------------------
B)
let x = adults
let y = children
8x+5y=4116
x+y=600
x=600-y
-----------------
8(600-y)+5y=4116
4800-8y+5y=4116
-3y=-684
y=228
----------------
x+y=600
x=600-228
x=372
----------------------
8*372+5*228=
2976+1140=4116
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
let x="gallons of 12%", so 10-x="gallons of 20%"

(12%)(x) + (20%)(10-x) = (14%)(10) ___ remember to change % to decimal

let x="number of adult tickets", so 600-x="number of children tickets"

(8)(x) + (5)(600-x) = 4116
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