My Math Forum Solving an inequation

 Algebra Pre-Algebra and Basic Algebra Math Forum

December 29th, 2017, 02:33 PM   #11
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Quote:
 Originally Posted by shaharhada Thanks. I want, please, a website that explain how to solve an inequation. Can I get a link?
https://www.mathsisfun.com/algebra/i...y-solving.html

I really don't know why you can't google these for yourself.

 December 29th, 2017, 03:08 PM #12 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,115 Thanks: 913 He's too busy...
December 29th, 2017, 07:24 PM   #13
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From: Nouakchott, Mauritania

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Math Focus: Algebra, Cryptography
Good morning !

Quote:
 Originally Posted by shaharhada Still I want a website that will explain me the gap in my learning of inequation. I don't learn this matter and I have a big PROBLEM because of that. Thanks. And google do not give me the links I needs.
I believe that all you need is not a whole website, but a general rule that can solve any kind of these inequalities which contain the absolute value.

So, let's take this simple but powerful rule

Rule :
Let $\displaystyle x$ be a real number and $\displaystyle a$ a positive real number.
Then :
$\displaystyle |x|<a\iff -a<x<a.$
This rule holds if we replace $\displaystyle <$ by $\displaystyle \le$.

For example : let's solve this inequality : $\displaystyle |2x-4|<8$.
We use the role above :
$\displaystyle -8<2x-4<8$
$\displaystyle -8+4<2x-4+4<8+4$
So :
$\displaystyle -4<2x<12$
And finally we divide by 2 :
$\displaystyle \frac{-4}2<\frac{2x}2<\frac{12}2$
We get :
$\displaystyle -2<x<6$.

It is not hard, is it ?

I hope that you can do the same with your question and - why not - all other inequality in your homework !

Have a good day !

 December 30th, 2017, 03:03 AM #14 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Do you understand what |x| means? If $\displaystyle 3x- 4\ge 0$ then $\displaystyle |3x- 4|= 3x- 4< 7$. $\displaystyle 3x< 7+ 4= 11$, $\displaystyle x< \frac{11}{3}$. But then we need to check if $\displaystyle 3x- 4\ge 0$ for x= 11/3. $\displaystyle 3x= 3(11/3)- 4= 11- 4= 7> 0$. Yes, $\displaystyle x> \frac{11}{3}$ satisfies this problem. If $\displaystyle 3x- 4< 0$ then $\displaystyle |3x- 4|= -(3x- 4)= -3x+ 4< 7$. [math-3x< 7- 4= 3[/math], $\displaystyle x> 3/3= 1$. Again, check: if x= 1 3x- 4= 3- 4= -1< 0. The solution set is "all x such than x< 11/3 or x> 1: 1< x< 11/3".
 December 30th, 2017, 03:31 AM #15 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,854 Thanks: 1078 Math Focus: Elementary mathematics and beyond That's incorrect.
 December 30th, 2017, 04:00 PM #16 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 608 Thanks: 82 0 is a counterexample to Country Boy. x = 0 makes the true equation 4 < 7. I would take Country Boy's -3x + 4 < 7, subtract 4 from both sides, and then solve -3x < 3 where you have to remember how to divide inequalities with negatives.

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