1. ## Solving inequalities

$\displaystyle \dfrac{m}{6}-5>-6$

I really forgot how to do this... Should i make that -5 a fraction?

2. You can do although I would simply multiply through each term by 6. Since 6 > 0 the direction of the inequality remains unchanged

$\displaystyle m - 30 > -36$

3. Originally Posted by vaironxxrd
$\displaystyle \dfrac{m}{6}-5>-6$

I really forgot how to do this... Should i make that -5 a fraction?
Or Add 5 on both sides of the inequality:

$\displaystyle \dfrac{m}{6}-5+5>-6+5$

$\displaystyle \dfrac{m}{6} > -1$

then multiply by 6

4. ## trying to solve it

I really don't know how that one works but here its what i have in mind

[Math]\dfrac{m}{6}-\dfrac{5}{1}>6[/tex] ?
I really don't know what i am doing

5. Originally Posted by harish21
Or Add 6 on both sides of the inequality:

$\displaystyle \dfrac{m}{6}-5+5>-6+5$

$\displaystyle \dfrac{m}{6} > -1$

then multiply by 6
That guy said multiply, and now you said add it, it looks like 3 different ways to do it now that made me
really confused

6. Originally Posted by harish21
Or Add 6 on both sides of the inequality:

$\displaystyle \dfrac{m}{6}-5+5>-6+5$

$\displaystyle \dfrac{m}{6} > -1$

then multiply by 6
by the way i multiply it is it m= -6?

7. Originally Posted by vaironxxrd
That guy said multiply, and now you said add it, it looks like 3 different ways to do it now that made me
really confused
e^(i*pi) said to multiply by 6 to clear the fraction

harish21 said to add 5 to isolate the variable term

they showed you two different methods to approach the problem

by the way i multiply it is it m= -6?
m > -6 ... it's an inequality, not an equation.

8. Originally Posted by skeeter
e^(i*pi) said to multiply by 6 to clear the fraction

harish21 said to add 5 to isolate the variable term

they showed you two different methods to approach the problem

m > -6 ... it's an inequality, not an equation.
Right sorry for that i got that on my paper forgot to put it.
i did another one just to prove im right ... is

5+$\displaystyle \dfrac{N}{2}$>8
how i solved this....

did -5 on the 8 got

$\displaystyle \dfrac{N}{2}$*$\displaystyle \dfrac{3}{1}$
N>6

RIGHT?!

9. $\displaystyle N > 6$ is correct but you have put multiply (*) in your working rather than greater than (>)

10. but it is already there in the (N/2)* (3/1)

.. Sorry if this is an off topic question but i got two more question for exponents and order of operations.. it has like 10 5 problems and im just asking for one in each section because i don't remember some of this stuff which is .... very simple

So should i create 1 more post with those 2 question which are different Topics but still under algebra category? or 2 different post?

11. Originally Posted by vaironxxrd
Right sorry for that i got that on my paper forgot to put it.
i did another one just to prove im right ... is

5+$\displaystyle \dfrac{N}{2}$>8
how i solved this....

did -5 on the 8 got

$\displaystyle \dfrac{N}{2}$*$\displaystyle \dfrac{3}{1}$
N>6

RIGHT?!
If that middle step was $\displaystyle \displaystyle \frac{N}{2} > 3$ then yes, it's correct.

12. Originally Posted by Prove It
If that middle step was $\displaystyle \displaystyle \frac{N}{2} > 3$ then yes, it's correct.
yes exactly what i meant.. I just made that 3 a fraction

13. Originally Posted by vaironxxrd
yes exactly what i meant.. I just made that 3 a fraction
That's not what I was pointing out, I was pointing out the use of a multiplication sign where the inequality sign should have been...

14. Originally Posted by Prove It
That's not what I was pointing out, I was pointing out the use of a multiplication sign where the inequality sign should have been...
I also did that earlier with a (*)