Santino Bautista

2022-06-27

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

Daniel Valdez

Beginner2022-06-28Added 19 answers

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

$x+3\ne 0\Rightarrow x\ne -3\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ 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 $(x+3)\cdot x$ , 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

$\frac{x-2}{x+3}\cdot \frac{x}{x}<\frac{x+1}{x}\cdot \frac{x+3}{x+3}$

$\frac{x(x-2)}{x(x+3)}<\frac{(x+1)(x+3)}{x(x+3)}$

This is equivalent to

$x(x-2)<(x+1)(x+3)$

Expand the parantheses to get

${\overline{){{x}^{2}}}}-2x<{\overline{){{x}^{2}}}}+x+3x+3$

Rearrange the inequality to isolate x on one side

$-6x<3\Rightarrow x>\frac{3}{(-6)}\iff x>-\frac{1}{2}$

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 $x\in (-\frac{1}{2},+\infty )\text{}\left\{0\right\}$

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

$x+3\ne 0\Rightarrow x\ne -3\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}$ 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 $(x+3)\cdot x$ , 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

$\frac{x-2}{x+3}\cdot \frac{x}{x}<\frac{x+1}{x}\cdot \frac{x+3}{x+3}$

$\frac{x(x-2)}{x(x+3)}<\frac{(x+1)(x+3)}{x(x+3)}$

This is equivalent to

$x(x-2)<(x+1)(x+3)$

Expand the parantheses to get

${\overline{){{x}^{2}}}}-2x<{\overline{){{x}^{2}}}}+x+3x+3$

Rearrange the inequality to isolate x on one side

$-6x<3\Rightarrow x>\frac{3}{(-6)}\iff x>-\frac{1}{2}$

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 $x\in (-\frac{1}{2},+\infty )\text{}\left\{0\right\}$

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