# Solve an inequality with fractions [closed]

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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## closed as not a real question by Bill Cook, Hans Lundmark, Kannappan Sampath, AD., Asaf KaragilaFeb 16 '12 at 0:19

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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
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
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
@JM: I've fixed it in the way it should be (I guess) – Ilya Feb 15 '12 at 16:08

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: $$21\times 2\quad ?\quad 4\times 5.$$
@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
@user1022734: sorry, I didn't get you. Do you know how to simplify $24/4$? – Ilya Feb 17 '12 at 11:11