## moongazer 4 years ago Teach me in solving this inequality.

1. moongazer

|2x−11|−2>4x+7

2. moongazer

I know how to solve equations. But I just don't know how to solve if it has absolute value sign. Teach me please. :)

3. anonymous

x < 1/3

4. moongazer

how?

5. moongazer

Teach me please and give me another example after teaching it. so that I can practice. and you will going to check it if I am right.

6. anonymous

To remove the modulus sign we write $|2x -11|$ as $\sqrt{2x - 11}^2$

7. moongazer

Why is that?

8. vishal_kothari
9. anonymous

To remove any modulus sign write $\left| ax + b \right| = (\sqrt{ax + b})^2$

10. moongazer

Could you solve it step by step? and what is the reason behind To remove any modulus sign write

11. moongazer

$\left| ax + b \right| = (\sqrt{ax + b})^2$

12. hoblos

|2x−11|−2>4x+7 |2x−11| > 4x +9 then 2x-11 > 4x +9 OR 2x-11 < -(4x+9) -2x > 20 2x -11 < -4x -9 x<-10 6x < 2 x<1/3

13. moongazer

14. hoblos

rules: if |x| < a then -a<x<a if |x| >a then x>a OR x<-a

15. hoblos

you apply the second rule here

16. moongazer

Thanks!

17. moongazer

What if it is "="

18. hoblos

if |x| = a then x=a OR x=-a