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anonymous
 5 years ago
73b= 5b+15
anonymous
 5 years ago
73b= 5b+15

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use the definition of absolute value; that is, the absolute value of a quantity, y, is defined by,\[y=\sqrt{y^2}\]So, for your problem, we have,\[\sqrt{(73b)^2}=\sqrt{(5b15)^2}\]You can then square both sides to get\[(73b)^2=(5b15)^2\]Expand and collect your terms into a quadratic equation\[16b^2108b+176=0\]and solve for b:\[b=4, \frac{11}{4}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am trying to find the possible values for the inequality in this problem.....i didnt think it was a quadratic equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You posted an equality...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The solutions are b = 1 and b = 11. Lokisan put 5b  15 instead 5b + 15 which led to incorrect answers. I did it by breaking the absolute value into separate equations. I'll do it this way: 7  3b = 5b + 15 > 7  3b = 5b + 15 OR 7  3b = 5b + 15. Equation A: 7  3b = 5b + 15 5b + 15 = 7  3b > 8b = 8 > b = 1 5b + 15 = (7  3b) = 3b  7 > 2b = 22 > b = 11 Equation B: 7  3b = 5b + 15 OR 3b  7 = 5b + 15 5b + 15 = 3b  7 > 2b = 22 > b = 11 5b + 15 = 7  3b > 8b = 8 > b = 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wow....thats a long problem lol but i will try it, thank you both

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and i was also thinking I had to set them both to zero because I had an example that looked the same that did that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's best to use either the definition of absolute value (like I did) or branlegr's case method.
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