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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.2012-02-15
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    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, $\frac{1}{4} = \frac{2}{x}$?2012-02-15
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    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)](http://en.wikipedia.org/wiki/Phrases_from_The_Hitchhiker%27s_Guide_to_the_Galaxy#Answer_to_the_Ultimate_Question_of_Life.2C_the_Universe.2C_and_Everything_.2842.29)2012-02-15
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    May be the editing was wrong?2012-02-15
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    I should ' multiplay over-cross ' to see if the equation is correct..bur how?2012-02-15
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    @Julian: [see for yourself.](http://math.stackexchange.com/posts/109651/revisions)2012-02-15
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    "Cross multiplication", you have mentioned; so you're **comparing** fractions if they're equal or not, not solving them...2012-02-15
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    YEs, sorry. English is not my main lang..2012-02-15
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    @JM: I've fixed it in the way it should be (I guess)2012-02-15
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    Thanks alot, but i solved it. :)2012-02-15

1 Answers 1

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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 then for any positive $k$ it holds that $ka 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 = 8/42012-02-15
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    @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 side2012-02-15
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    4x = 24 , so x = 6, but can i say x = 6 or x = 24/4 ? or it dosent matter?2012-02-16
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    @user1022734: sorry, I didn't get you. Do you know how to simplify $24/4$?2012-02-17