# math

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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 -

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 -

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 break-even number.

• math -

hi I THINK THE ANSWER IS 569

• math -

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 -

12 smalls 6 larges

• math -

12 small 6 large