how do i solve inequalities with unequal factors?

Division? What does the "unequal"ness of the factors have to do with anything?

The constant is attached to the variable by multiplication. Use division to get it off. This ides does not change based on any "factors".

Note on INequalities. Multiplication or division by negative numbers requires the reversal of the inequality. You can prove this by addition.

-x < 3

By division (-1) gives x > -3

By addition (x) and subtraction (-3) gives -3 < x
 
inequalities

My daughter has a new issue, the teacher is not explaining how to do a fraction inequality. Please help!! 3/4m < or equal to 1/4m+2[/u]
 
You are doing it again. Why are there so many versions of inequalities. They are ALL the same. Fractions are numbers. Treat them like any other number. Equal or unequal, it is of no consequence. Ugly, cure, pretty, or plastic, it is the same.

Integers - Unequal:

3x < 6

Attachment is multiplication. Use division

3x/3 < 6/3

x < 2

Integers - Equal

6x < 6

Attachment is multiplication. Use division

3x/6 < 6/6

x < 1

Fractions

(1/3)x < 6

Attachment is multiplication. Use division

(1/3)x/(1/3) < 6/(1/3)

x < 18

Animals

(Frog)x < (Cat)

Attachment is multiplication. Use division

(Frog)x/(Frog) < (Cat)/(Frog)

x < (Cat)/(Frog)

Inanimate Objects

(House)x < (Pizza)

Attachment is multiplication. Use division

(House)x/(House) < (Pizza)/(House)

x < (Pizza)/(House)

It is ALL the same. Do NOT confuse yourself with some concept that each new problem is a different experience.
 
jamiej said:
My daughter has a new issue, the teacher is not explaining how to do a fraction inequality. Please help!! 3/4m < or equal to 1/4m+2[/u]
Click to expand...
To help your daughter, it would be best if we could hear from her. Trying to tutor through a "translator" who doesn't "speak the language" has, historically, very little chance of success. Sorry! :oops:

Please have your daughter reply, or dictate a reply, indicating whether she is comfortable solving linear equations, regardless of the presence of fractions. If so, then she's pretty much there. If not, then we need to provide her with links to lessons on fractions, so that she can become comfortable with them. Then she can get comfortable with fractions in the simpler context of linear equations. And then we can move on to linear inequalities. :idea:

Thank you! :D

Eliz.
 
I don't want to learn how to do linear equations or linear inequalities I just want to know how to solve two-step equations with frctions(on both sides). :) :?
 
jamiej said:
I don't want to learn how to do linear equations or linear inequalities I just want to know how to solve two-step equations with frctions
Click to expand...
Your position is to be regretted, since the methods for solving linear equations (or inequalities) with fractions are exactly the same methods for solving linear equations (or inequalities) without fractions. One learns the easier case (without) and then moves on to the harder one (with).

Since you don't want to learn, and since legitimate tutors don't "do" students' work for them, it would appear that we will be unable to help you.

My best wishes to you.

Eliz.
 
sorry for the misunderstanding

i didn't know that those were the same because my teacher has not taught us that yet! i'm very lost so please help. I do not understand how to divide fractions and how to divide fractions by whole numbers.

my teacher gave me an assignment that started with these kind of problems:
5a +6 > -9
-6 -6
5a/5 > -15/5= a > -5
further into the assignment i got these kind of problems:
-1/5x > 4/5x +3
how do I solve this?
 
it didn't really help, what do I do first in this problem: 3/4m <1/4m +2
 
jamiej said:
it didn't really help, what do I do first in this problem: 3/4m <1/4m +2
Click to expand...

Isolate 'm'

samething you would do - if you had to solve,

3m = 4m - 2
 
Do you mean:
\(\displaystyle \frac{3}{4m} < \frac{1}{4m} + 2\) or \(\displaystyle \frac{3}{4m} < \frac{1}{4m + 2}\)?

My hint to you: Treat it as if the < sign was an equal sign. The only difference is that when you divide by a negative number or take the reciprocal, you would have to switch signs (from < to > or from > to <):

ex.) -4m < 8 \(\displaystyle \to\) 4m > -8 (Divided both sides by -1)
\(\displaystyle \frac{5}{m} < 2 \to \frac{m}{5} > \frac{1}{2}\) (Took the reciprocal of both sides)

Otherwise, you would take the same procedures when dealing with these kinds of inequalities as if you were dealing with an equal sign.
 
I think the problem that the student trying to solve is:

\(\displaystyle \frac{3}{4}\cdot{m} < \frac{1}{4}\cdot{m} + 2\)
 
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