# Math Help - Two problems with solutions...help!!

1. ## Two problems with solutions...help!!

Sergio plans to paint the 4 walls of his room with 1 coat of paint. The walls are rectangular, and, according to his measurements, each wall is 10ft by 15ft. He will not need to paint the single 3-foot-by-5-foot rectangular window in his room and the 3.5 foot-by-7-foot rectangular door. Sergio knows that each gallon of paint covers between 300 and 350 square feet. If only 1-gallon cans of paint are available, what is the minimum number of cans of paint Sergio needs to buy to paint his walls?

The book says the solution is 2 but I can't figure it out??

Also, I need to solve the following:

8^(2x + 1) = 4^(1 - x)

The solution is -1/8...not sure why though...any help would be appreciated on these.

2. There are 4 10x15 ft walls. A walls measures are 10x15 ft and its area is 150 ft². 4 walls will make 150x4 = 600 ft². As he won't paint the 3x5 ft (15 ft²) window and the 3.5x7 ft (24.5 ft²) door, we can subtract them from the total area of walls.
600 - 15 = 585
585 - 24.5 = 560.5 ft²

If 1 gallon can of paint can cover 300-350 ft², and the question asks for the best situation, we will take it as 350 ft². So he will need 2 cans to paint 560.5 ft².

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$8^{2x + 1} = 4^{1 - x}$

$(2^3)^{2x+1} = (2^2)^{1-x}$

$2^{6x+3} = 2^{2-2x}$

$6x + 3 = 2 - 2x$

$8x = -1$

$x = - \frac{1}{8}$

3. Thanks so much wingless--can't believe I didn't think of that for both problems!

4. You're welcome. Don't worry, as many questions as you solve, as much as you think on them, you'll get better at 'seeing things' in questions