Inequality with unknown in denominator

[stickychar]

I just cannot seem to do this problem. The logical steps are not working and I dont understand why?
It is as follows:
(3-2x)/x </= -1

what i thought to do is this.
3-2x</=-x
3</=x
it is incorrect, help would be appreciated, thanks, charlie

 

[daon2]

You have only half of the solution. In your work, you presume that x is not a negative number. So what if x<0?

 

[soroban]

Hello, stickychar!

$\displaystyle \dfrac{3-2x}{x} \:\le \:-1$

What i thought to do is this:. $\displaystyle 3-2x\:\le\: -x$
This is a careless step.

Your step is correct if $\displaystyle x$ is positive.

If $\displaystyle x$ is negative, you are multiplying by a negative quantity:
. . the inequality is reversed.
Then we would have: .$\displaystyle 3-2x \:\ge \: -x$

You must consider both cases.


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If the variable is in the numerator and denominator,
. . this situation will arise.

There is another way to handle it . . .

Bring all terms to one side: .$\displaystyle \dfrac{3-2x}{x} + 1 \:\le\:0$

Combine terms: .$\displaystyle \dfrac{3-x}{x} \:\le\:0$


Consider what the statement says: a fraction is negative.
This is true if the numerator and denominator have opposite signs.

Now examine the two cases . . .

(1) Positive numerator, negative denominator

. . . $\displaystyle \begin{array}{cccccccccc}3-x \:\ge\:0 & \Rightarrow & 3\:\ge\: x \\ x \:<\:0 \end{array}$

We have: .$\displaystyle \begin{Bmatrix}x\:\le\:3 \\ x\:<\:0\end{Bmatrix}$

Both inequalties are satisfied if: .$\displaystyle \boxed{x \:<\:0}$
. . . (We take the "stronger" inequality.)


(2) Negative numerator, positive denominator

. . . $\displaystyle \begin{array}{cccccc}3-x \:\le\:0 & \Rightarrow & 3 \:\le\: x \\ x \:>\:0 \end{array}$

We have: .$\displaystyle \begin{Bmatrix} x\:\ge\:3 \\ x\:>\:0 \end{Bmatrix}$

Both inequalities are satisfied if: .$\displaystyle \boxed{x\:\ge\:3}$


$\displaystyle \text{The solution is: }\;x\:<\:0\;\text{ or }\; x\:\ge\:3$

$\displaystyle \text{Interval notation: }\;(-\infty,\,0) \,\cup\,[3,\,\infty)$

$\displaystyle \begin{array}{ccccccc}\text{Graph: }& === & \circ & --- & \bullet & === \\ && 0 && 3 \end{array}$