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How can an inequality with fractions should be solved ?

Let say :

$$ \displaystyle \frac{2}{4}\quad?\quad\frac{5}{21} $$

Please give me examples, information (step by step).

I should multiply over-cross ' to see if the equation is correct

--------------------------------------------------------- > Solved

Used: a/b = c/d => a*d = b*c

44 = 20 , becouse it not the same on both side, the fraction is wrong.

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You do realize $2\cdot 21 \neq 5\cdot 4$? and that what you have is not an equation. Re-write the answer by editing it please. –  Pedro Tamaroff Feb 15 '12 at 15:54
1  
What are we solving? You've written down two fractions that are unequal and placed "$=$" between them. Do you want to solve for $x$ in something like, say, $\frac{1}{4} = \frac{2}{x}$? –  Dylan Moreland Feb 15 '12 at 15:56
4  
First of all, if $a = b$ if and only if $ka=kb$ for all $k\neq 0$. 1. we apply $k = 4$ and obtain $2 = \frac{20}{21}$; 2. we apply $k = 21$ and obtain $42 = 20$; since there is $42$ in the LHS, you've got the answer (to the Ultimate Question of Life, the Universe, and Everything) –  Ilya Feb 15 '12 at 15:56
    
May be the editing was wrong? –  Julian Kuelshammer Feb 15 '12 at 15:57
1  
@JM: I've fixed it in the way it should be (I guess) –  Ilya Feb 15 '12 at 16:08
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closed as not a real question by Bill Cook, Hans Lundmark, Kannappan Sampath, AD., Asaf Karagila Feb 16 '12 at 0:19

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1 Answer

up vote 6 down vote accepted

The main idea is given in my comment: of course, you can use a cross-multiplication to solve this inequality - but why does it work? There is an rule (which is an axiom for inequalities) that if $a<b$ then for any positive $k$ it holds that $ka<kb$ and for any negative $l$ it holds that $la>lb$.

Let us consider your example, you have $$ \displaystyle{\frac 24 \quad?\quad \frac{5}{21}}. $$ Whatever sign $?$ denotes, if we multiply both sides by a positive number, the sign does not change. So we multiply both sides by both denominators and obtain $$ 21\times 4\times \frac24 \quad?\quad 4\times 21\times\frac{5}{21} $$ and hence $$ 42\quad?\quad20 $$ so $?$ is $>$.

Then what about the cross-multiplication? You do the same but you write instead $$ 21\times \left(4\times \frac24\right) \quad?\quad 4\times \left(21\times\frac{5}{21}\right) $$ and since the denominators cancel it is equivalent to the cross-multiplication rule: $$ 21\times 2\quad ?\quad 4\times 5. $$

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Good, but how can you solve fraction equation with cross-multiplication ? let say: x/3 = 8/4 –  user1022734 Feb 15 '12 at 16:24
    
@user1022734: If you solve it by cross-multiplication then you obtain $4x = 3\times 8$, so then you have to divide again. If you just follow the method I've described - you just need to multiply both sides with $3$ in order to obtain a single $x$ in the left hand side –  Ilya Feb 15 '12 at 16:27
    
4x = 24 , so x = 6, but can i say x = 6 or x = 24/4 ? or it dosent matter? –  user1022734 Feb 16 '12 at 19:19
    
@user1022734: sorry, I didn't get you. Do you know how to simplify $24/4$? –  Ilya Feb 17 '12 at 11:11
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