Solving inequalities

**vaironxxrd** · February 22nd 2011, 04:15 PM

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

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

**e^(i*pi)** · February 22nd 2011, 04:18 PM

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

$m - 30 > -36$

**harish21** · February 22nd 2011, 04:20 PM

Originally Posted by vaironxxrd

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

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

Or Add 5 on both sides of the inequality:

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

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

then multiply by 6

**vaironxxrd** · February 22nd 2011, 04:24 PM

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

**vaironxxrd** · February 22nd 2011, 04:26 PM

Originally Posted by harish21

Or Add 6 on both sides of the inequality:

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

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

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

**vaironxxrd** · February 22nd 2011, 04:29 PM

Originally Posted by harish21

Or Add 6 on both sides of the inequality:

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

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

then multiply by 6

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

**skeeter** · February 22nd 2011, 04:35 PM

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.

**vaironxxrd** · February 22nd 2011, 04:41 PM

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

did -5 on the 8 got

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

RIGHT?!

**e^(i*pi)** · February 22nd 2011, 04:46 PM

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

**vaironxxrd** · February 22nd 2011, 04:49 PM

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?

**Prove It** · February 22nd 2011, 04:53 PM

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

did -5 on the 8 got

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

RIGHT?!

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

**vaironxxrd** · February 22nd 2011, 04:58 PM

Originally Posted by Prove It

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

yes exactly what i meant.. I just made that 3 a fraction

**Prove It** · February 22nd 2011, 05:09 PM

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...

**vaironxxrd** · February 22nd 2011, 05:21 PM

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 (*)

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