## anonymous 5 years ago |7-3b|= |5b+15|

1. anonymous

Use 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,$\sqrt{(7-3b)^2}=\sqrt{(5b-15)^2}$You can then square both sides to get$(7-3b)^2=(5b-15)^2$Expand and collect your terms into a quadratic equation$16b^2-108b+176=0$and solve for b:$b=4, \frac{11}{4}$

2. anonymous

i am trying to find the possible values for the inequality in this problem.....i didnt think it was a quadratic equation.

3. anonymous

You posted an equality...

4. anonymous

The 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

5. anonymous

6. anonymous

oh wow....thats a long problem lol but i will try it, thank you both

7. anonymous

and i was also thinking I had to set them both to zero because I had an example that looked the same that did that

8. anonymous

It's best to use either the definition of absolute value (like I did) or branlegr's case method.