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This should be something relatively simple. I know there's a trick to this, I just can't remember it. I have an equation $$\frac{3x}{x-3}\geq 4.$$ I remember being shown at some point in my life that you could could multiply the entire equation by $(x-3)^2$ in order to get rid of the divisor. However, $$3x(x-3)\geq 4(x-3)^2.$$ Really doesn't seem to go anywhere.

Anyone have any clue how to do this, that could perhaps show me/ point me in the right direction?

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Try subtracting the right-hand side from the left-hand side in your final inequality. –  Yuval Filmus Mar 7 '11 at 16:14
    
Please use parentheses to show what you mean. 3x/x-3 should divide before subtracting and result in 0. –  Ross Millikan Mar 7 '11 at 16:27
    
In my answer here, I describe the "boundary algorithm" for solving such inequalities (both the original one and the result of the multiplication). –  Isaac Mar 8 '11 at 9:25

3 Answers 3

up vote 6 down vote accepted

That method will work, but there's actually a simpler more general way. But first let's finish that method. $\:$ After multiplying through by $\rm\: (x-3)^2\: $ (squared to preserve the sense of the inequality) we obtain $\rm\ 3\:x\:(x-3) \ge\: 4\:(x-3)^2\:.\:$ Putting all terms on one side and factoring out $\rm\ x-3\ $ we obtain $\rm\: p(x) = (x-3)\:(a\:x-b) \ge 0\ $ for some $\rm\:a\:,\:b\:.\:$ The (at most) two roots partition the real line into (at most) three intervals where $\rm\ p(x)\ $ has constant sign. So the answer follows by simply testing these few intervals to determine where $\rm\:p(x)\:$ has the desired sign.

But multiplying through by $\rm\: (x-3)^2\: $ creates more work than necessary. Instead we can simply multiply through by $\rm\: x-3\: $ and worry about the signs later, since we really only need to know the other root (besides $\rm\ x = 3\:$) of the quadratic $\rm\: p(x)\:.\:$ Indeed, if we multiply by $\rm\ x-3\ $ we get either $\rm\ 3x \ge 4\:x-12\ \Rightarrow\ 12 \ge x\ $ or we get the reverse $\rm\ 12 \le x\:,\:$ depending on the sign of $\rm\ x-3\:.\: $ Either way we get the same root $\rm\: x = 12\:,\: $ so we obtain the same partition of the real line into intervals where the function has constant sign. This method is simpler since one works with lower degree polynomials, here linear vs. quadratic in the first method above.

Similar remarks hold true for arbitrary rational functions. Suppose that $\rm\: A,B,C,D\: $ are polynomials in $\rm\:x\:$ and we seek to determine where $\rm\: A/B\ >\ C/D\:.\: $ The first method converts to polynomials by multiplying it by $\rm\: (BD)^2\: $ resulting in $\rm\: ABD^2 >\: CDB^2\ $ or $\rm\ BD\ (AD-BC) > 0\:.\:$ For the second method, we partition the real line by the roots of the denominators $\rm\: B,D\ $ and also by the roots of the polynomial obtained by multiplying through by $\rm\:BD\:,\:$ i.e. the roots of $\rm\: AD-BC\:.\:$ However the union of these root sets is precisely the same as the roots of $\rm\: BD\ (AD-BC)\: $ from the first method. Since both methods partition the real line into the same constant-sign intervals, they're equivalent.

If you study rational (and algebraic) functions and their Riemann surfaces you'll see that this is just a special case of how their roots (including poles = roots at $\infty$) serve to characterize them. From another viewpoint, the above method can be viewed as a very special case of Tarski's quantifier elimination algorithm for deciding arbitrary real polynomial inequalities - which, like above, works by decomposing $\rm\: \mathbb R^n\: $ into a finite number constant-sign (semi-)algebraic "cylinders".

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How cool is that! Thx. –  Alex Mar 9 '11 at 6:53
    
Stupid question, but how does one plot the points once you've got x=12? –  Alex Mar 9 '11 at 7:07
1  
@Alex: You can use Wolfram Alpha to plot it. If you mean to ask how to do it by hand then it would be best to ask another question on that topic. –  Bill Dubuque Mar 9 '11 at 7:32

I'll give a more mechanical way to solve the inequality that avoids the (neat!) trick of multiplying by $(x-3)^2$. But note that this is essentially the same as what Bill and Eivind showed. From $$ \frac{3x}{x-3} \geq 4 , $$ we get $$ \frac{3x}{x-3} - 4 \geq 0 \hspace{.2in} \implies \frac{3x-4x+12}{x-3} \geq 0. $$ Simplifying this, we get $$ \frac{12-x}{x-3} \geq 0. $$

From this point, the solution is similar to Eivind's. This inequality implies one of the following two cases:

  • $12 - x \geq 0$ and $x-3 > 0$. That is, $3 < x \leq 12$.

  • $12 - x \leq 0$ and $x - 3 < 0$. That is, $x < 3$ and $x \geq 12$; clearly this is impossible.

Hence the valid solutions are $3 < x \leq 12$.

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$$\begin{align} 3x(x-3)&\geq 4(x-3)^2 \\ 3x^2-9x &\geq 4(x^2-6x+9) \\ 3x^2-9x &\geq 4x^2-24x+36 \end{align}$$ By rearranging we get $$\begin{align} x^2-15x+36 &\leq 0 \\ (x-3)(x-12) &\leq 0 \end{align} $$ So $(x-3)> 0 \land (x-12)\leq0$ or $(x-3)< 0 \land (x-12)\geq 0$. ($x-3\neq0$)

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I think it would be best to note that $x-3 \ne 0$, or the original equation would be undefined at $x = 3$. –  Eugene Bulkin Mar 8 '11 at 4:32
    
Yes, thank you. –  please delete me Mar 8 '11 at 8:26

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