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(x-2)/(x+3) > -2

$\frac{x-2}{x+3}>-2$

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Solve advanced linear inequalities

Full solution

Q.

\frac{x-2}{x+3}>-2

Eliminate Fraction by Multiplication: First, we need to get rid of the fraction by multiplying both sides of the inequality by $(x+3)$ , which is the denominator. However, we must consider that the inequality sign will change direction if we multiply by a negative number. Since we don't know if $(x+3)$ is positive or negative, we will have to consider two cases: one where $(x+3)$ is positive and one where $(x+3)$ is negative.
Case $1$ : $x > -3$ : Case $1$ : Assume $(x+3) > 0$ , which means $x > -3$ . Now multiply both sides of the inequality by $(x+3)$ to eliminate the fraction. $\newline$ $\frac{x-2}{x+3} \cdot (x+3) > -2 \cdot (x+3)$ $\newline$ $x - 2 > -2x - 6$
Combine Terms and Solve: Now, we will combine like terms and solve for $x$ . $x + 2x > -6 + 2$ $3x > -4$ $x > -\frac{4}{3}$
Case $2$ : $x < -3$ : Case $2$ : Assume $(x+3) < 0$ , which means $x < -3$ . When we multiply both sides by a negative number, we must flip the inequality sign. $\newline$ $\frac{x-2}{x+3} \cdot (x+3) < -2 \cdot (x+3)$ $\newline$ $x - 2 < -2x - 6$
Combine Terms and Solve: Combine like terms and solve for $x$ in this case. $\newline$ $x + 2x < -6 + 2$ $\newline$ $3x < -4$ $\newline$ $x < -\frac{4}{3}$ $\newline$ However, this result contradicts our assumption that $x < -3$ . Since $-\frac{4}{3}$ is greater than $-3$ , we cannot include this in our solution set.
Combine Results from Both Cases: Now we need to combine the results from both cases. From Case $1$ , we have $x > -\frac{4}{3}$ , and this is valid when $x > -3$ . Therefore, the solution set for the inequality is $x > -\frac{4}{3}$ , but we must exclude $x = -3$ because the original inequality is undefined at $x = -3$ .

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