I have the inequality:

$$(14x)/(5-x) \le 7x$$

I tried solving it this way: $$ 14x \le 7x(5-x) \\ 2 \le (5-x) \\ -3 \le -x \\ x \le 3 $$

Apparently, the answer is $$0\le x\le 3$$

What am I doing wrong?


You multiplied by $5-x$ and you divided by $7x$. Not only is the latter impossible for $x=0$ but for $5-x,x<0$ the sign of the inequality would have flipped. The correct route is to first seperate the problem into two cases. Either $x>5$ or $x<5$ (can't have $x=5$) or the LHS won't exist.

In the first case, $5-x<0$ and therefore the equality becomes

\begin{equation*}\begin{aligned}14x &\geq 7x(5-x)\\ 2x &\geq 5x-x^2\\ x^2-3x &\geq 0\\ x(x-3)&\geq 0\end{aligned}\end{equation*}

which holds only if $x\geq 3$ or $x\leq 0$. But we are already assuming that $x>5$, so in this case only $x>5$ is a valid solution (we discard the other values).

In the second case, $5-x>0$ and therefore the equality becomes

\begin{equation*}\begin{aligned}14x &\leq 7x(5-x)\\ 2x &\leq 5x-x^2\\ x^2-3x &\leq 0\\ x(x-3)&\leq 0\end{aligned}\end{equation*}

and this last equation can only hold between the two roots of the quadratic i.e. $0\leq x\leq 3$. This time we're assuming that $x<5$ so this makes no difference to our solution.

Therefore the overall solution in $0\leq x \leq 3$ and $x>5$. Your teacher was wrong.

  • $\begingroup$ In the first step, you multiplied both sides by (5-x). Is this allowed? Since (5-x) can be negative as well. What would be the best way to approach this sort of problems? $\endgroup$ – juan981 May 18 '14 at 22:27
  • $\begingroup$ Oh yeah, that's naughty too... $\endgroup$ – Rhidian May 18 '14 at 22:30
  • $\begingroup$ @Rhidian: you missed the case x > 5 $\endgroup$ – DeepSea May 18 '14 at 22:36
  • $\begingroup$ @LAcarguy Yeah, I didn't bother reading the first part of the OP's solution. Corrected now. $\endgroup$ – Rhidian May 18 '14 at 22:42

Error 1: $5 - x$ can be negative, so you can't multiply it both sides.

Error 2: $x$ can also be negative, so you can't remove it in your line #3.

Solution: $\dfrac{2x}{5-x} - x \leq 0 \iff \dfrac{2x - 5x + x^2}{5-x} \leq 0 \iff \dfrac{x(x-3)}{5-x} \leq 0$. There are $3$ zeros: $0$, $3$, and $5$ for both numerator and denominator. Notice $x \neq 5$, and pick a number to check the sign of the $LHS$. Choose $x = 4$, then $LHS = 4 > 0$. This means: $0 \leq x \leq 3$ or $x > 5$.


your process is fine, but don't have a under limit in the equation below x, then the only thing you should do is realize that you put the lower limit never exists therefore x can take values ​​for x <= 3


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