Question

Solve each inequality algebraically. $$\frac{x-1}{x+2} \geq-2$$

Short answer

The solution is \( (-\infty, -2) \cup [-1, \infty) \).

Step-by-step solution

Move the constant term to the other side

Start by isolating the fraction on one side of the inequality. Subtract \( -2 \) from both sides:\( \frac{x-1}{x+2} + 2 \geq 0 \)

Combine like terms

Convert the constant term into a fraction with the same denominator. This results in:\( \frac{x-1 + 2(x+2)}{x+2} \geq 0 \)

Simplify the numerator

Combine and simplify the terms in the numerator:\( \frac{x-1 + 2x + 4}{x+2} = \frac{3x + 3}{x+2} \)

Factor the numerator

Factor the numerator to simplify the expression further:\( \frac{3(x + 1)}{x+2} \geq 0 \)

Find critical points

Determine the values of \( x \) that make the numerator and denominator zero (critical points):From \(3(x + 1) = 0\), we get \(x = -1\). From \(x + 2 = 0\), we get \(x = -2\).

Test intervals

Test the sign of the expression in the intervals defined by the critical points: \((-\infty, -2)\), \((-2, -1)\), and \((-1, \infty)\).Choose test points from each interval to determine the sign of the expression.

Determine solution interval

Evaluate the sign of \( \frac{3(x + 1)}{x+2} \) in each interval:- For \(x < -2\), the expression is positive.- For \(-2 < x < -1\), the expression is negative.- For \(x > -1\), the expression is positive.Combine the intervals where the expression is non-negative. Also, include the points where the expression equals zero (\(x = -1\)).

Conclusion

Combine the results to find the solution set:The solution is \( (-\infty, -2) \cup [-1, \infty) \).

Key concepts

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations. In the inequality \(\frac{x-1}{x+2} \geq-2\), the algebraic expression is \(\frac{x-1}{x+2}\). It's important to manipulate these expressions correctly to isolate the variable.
First, we add 2 to both sides to move the constant: \( \frac{x-1}{x+2} + 2 \geq 0\).
This step transforms the inequality into a form that allows us to combine like terms and simplify the expression further.
By converting the constant 2 into a fraction with the same denominator, i.e., \(\frac{2(x+2)}{x+2}\), we can combine the algebraic expressions into \(\frac{x-1 + 2(x+2)}{x+2} \geq 0\).
This results in a new fraction with a simplified numerator. Algebraic manipulation is vital in solving inequalities, as it helps in isolating and reasoning through the variable.

Rational Inequalities

Rational inequalities involve expressions in the form of fractions, where the numerator and/or denominator has variables. To solve rational inequalities, such as \(\frac{x-1}{x+2} \geq-2\), we transform them into an equivalent inequality with a zero on one side.
We do this because it's easier to analyze the sign of a rational expression compared to comparing it directly to another number.
After the initial transformation and simplification steps, we get \(\frac{3(x+1)}{x+2} \geq 0\).
The inequality now involves understanding when the fraction is zero or positive (non-negative in this context). Rational inequalities are more complex than linear inequalities because we must consider both the numerator and the denominator.

Critical Points

Critical points in the context of inequalities are values of the variable that make the numerator or denominator zero, changing the sign of the inequality. These points help divide the number line into manageable intervals for analysis.
For \(\frac{3(x+1)}{x+2} \geq 0\), we find critical points by solving when the factors equal zero: \(3(x+1) = 0\) gives \(x = -1\) and \(x+2 = 0\) gives \(x = -2\).
These values must be tested separately as they can potentially switch the inequality from positive to negative or vice versa.
Identifying critical points is crucial because they determine the intervals we need to test to find the solution set of the inequality.

Interval Testing

Interval testing is the method of analyzing the sign of a rational expression within different intervals defined by the critical points. For the inequality \(\frac{3(x+1)}{x+2} \geq 0\), we break the number line into intervals based on the critical points: \((-\infty, -2)\), \((-2, -1)\), and \((-1, \infty)\).
We then choose test points from each interval to determine the sign of the expression within those intervals. For example:
- For \(x < -2\), the expression is positive.
- For \(-2 < x < -1\), the expression is negative.
- For \(x > -1\), the expression is positive.
This step helps to pinpoint where the inequality holds true. Finally, combining the positive intervals, including where the expression equals zero, gives us the solution set: \((-\infty, -2) \cup [-1, \infty)\).