# 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

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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