Posted by Jane on .
An artist is going to sell two sizes of prints at an art fair. The artist will charge $20 for the small print and $45 for a large print. The artist would like to sell twice as many small prints as large prints. The booth the artist is renting for the day cost $510. How many of each size print must the artist sell in order to break even at the fair?

math 
Reiny,
number of small prints  x
number of large prints  2x
20x + 45(2x) = 510
20x+90x=510
110x = 510
x= 510/110 = 4.6
but you can't sell .6 of a print, so I would say he should sell 5 small and 10 large prints
check:
5(20) + 10(45) = 550  a profit
if he sold only 4 small and 8 large
income = 4(20) = 8(45) = 440 , not enough to cover rent. 
math 
choochoo,
You switched the large and small print numbers. Selling twice as many small prints would make it:
20 (2x) + 45 (x) = 510
If you solve with that formula in mind, x = 6...and that is the breakeven number. 
math 
samuel,
hi I THINK THE ANSWER IS 569

math 
RR,
20x(2) + 45x = $510
40x + 45x =$510
85x / 85 = $510/85
x = 6
She sold twice as many small than large. So, small = 2(6) which equals 12
20(12) + 45(6) = $510
240 + 270 = $510 
math 
swagmath,
12 smalls 6 larges

math 
person,
12 small 6 large