Santino Bautista

2022-06-27

How do you solve $\frac{x-2}{x+3}<\frac{x+1}{x}$ ?

Daniel Valdez

Step 1
The inequality given to you looks like this
$\frac{x-2}{x+3}<\frac{x+1}{x}$
Right from the start, you know that any solution interval must not contain the values of x that will make the two denominators equal to zero.
More specifically, you need to have
$x+3\ne 0⇒x\ne -3\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ and $\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}x\ne 0$
With that in mind, use the common denominator of the two fractions, which is equal to $\left(x+3\right)\cdot x$ , to get rid of the denominators.
More specifically, multiply the first fraction by $1=\frac{x}{x}$ and the second fraction by
$1=\frac{x+3}{x+3}$
This will get you
$\frac{x-2}{x+3}\cdot \frac{x}{x}<\frac{x+1}{x}\cdot \frac{x+3}{x+3}$
$\frac{x\left(x-2\right)}{x\left(x+3\right)}<\frac{\left(x+1\right)\left(x+3\right)}{x\left(x+3\right)}$
This is equivalent to
$x\left(x-2\right)<\left(x+1\right)\left(x+3\right)$
Expand the parantheses to get
$\overline{){x}^{2}}-2x<\overline{){x}^{2}}+x+3x+3$
Rearrange the inequality to isolate x on one side
$-6x<3⇒x>\frac{3}{\left(-6\right)}⇔x>-\frac{1}{2}$
This means that any value of x that is greater than $-\frac{1}{2}$ , except $x=0$ , will be a solution to the original inequality.
Therefore, the solution interval will be $x\in \left(-\frac{1}{2},+\infty \right)\text{\}\left\{0\right\}$

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