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Solve the inequality $$\frac{2x-3}{x+4} \leq 0$$ Give your answer in interval notation.

$$(-4,3/2] \cup [3/2, \infty)$$

Am I correct?

  • 0
    What is the $$\lim_{x\to \infty}\frac{2x-3}{x+4}$$? – 2017-01-09
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    Could you include your work that got you to that answer? – 2017-01-09
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    Checking some values in the interval you suggest was not an option because? – 2017-01-10

3 Answers 3

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$\frac{2x-3}{x+4}\leq0\iff$

$[(x+4>0)\wedge(2x-3\leq0)]\vee[(x+4<0)\wedge(2x-3\geq0)]\iff$

$[(x>-4)\wedge(x\leq1.5)]\vee[(x<-4)\wedge(x\geq1.5)]\iff$

$(-4

$-4

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Hint: take a good look at the following plot of the graph of $f (x) := \dfrac{2x-3}{x+4}$.

enter image description here

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Hint: Note that your proposed interval \begin{align*} \left(\left.-4,3/2\right]\right.\cup\left[\left.3/2,\infty\right)\right.=\left(-4,\infty\right) \end{align*} is not the correct solution, since e.g. $2\in \left(-4,\infty\right)$, but $\left.\frac{2x-3}{x+4}\right|_{x=2}=\frac{1}{6}>0$

Range check: At first we check the range of definition of the inequality.

  • We observe the inequality is defined for all real values as long as the denominator of $\frac{2x-3}{x+4}$ is not equal $0$. We conclude the inequality is defined for all $x\in\mathbb{R}\setminus \{-4\}$.

Next we want to determine the set of $x\in\mathbb{R}\setminus \{-4\}$ where the inequality is true. In order to do so, we multiply with the denominator $x+4$.

We have to respect that multiplication of an inquality with a positive quantity preserves the inequality sign, while multiplication with a negative quantity inverts the inequality sign. So, we consider two cases:

Case: $x+4>0$

In case $x+4>0$ or equivalently $x>-4$ we obtain \begin{align*} \frac{2x-3}{x+4} &\leq 0\\ 2x-3 &\leq 0\cdot (x+4)\\ 2x-3 &\leq 0\\ 2x&\leq 3\\ x&\leq \frac{3}{2} \end{align*}

We see, the inequality is valid, if $-4< x\leq \frac{3}{2}$ resp. in interval notation \begin{align*} x\in(-4,3/2] \end{align*}

$$ $$

Case: $x+4<0$

In case $x+4<0$ or equivalently $x<-4$ we obtain \begin{align*} \frac{2x-3}{x+4} &\leq 0\\ 2x-3 &\geq 0\cdot (x+4)\\ 2x-3 &\geq 0\\ 2x&\geq 3\\ x&\geq \frac{3}{2} \end{align*}

We see, the inequality is never valid in this case, since there is no $x$ with $x<-4$ and at the same time $x\geq \frac{3}{2}$.

So, only the first case provides solutions for the inequality.

Conclusion: The inequality is true for $x\in \mathbb{R}\setminus \{-4\}$ if and only if \begin{align*} x\in(-4,3/2] \end{align*}