Find the set of real values of x that satisfies the inequality \frac{3 - x}{x + 2} \geq 1. — JAMB UTME

Find the set of real values of $x$ that satisfies the inequality $\frac{3 - x}{x + 2} \geq 1$ .

-2 < x \leq \frac{1}{2}

Subtracting 1 from both sides yields

\frac{1 - 2x}{x + 2} \geq 0

. The critical points are

x = \frac{1}{2}

and

x = -2

. Testing values shows that the fraction is positive for

-2 < x < \frac{1}{2}

and equal to zero at

x = \frac{1}{2}

. Since

x = -2

causes division by zero, it is excluded from the interval, giving

-2 < x \leq \frac{1}{2}

Subtract 1 from both sides of the inequality to set one side to zero.

\frac{3 - x}{x + 2} - 1 \geq 0

Direct cross-multiplication is invalid because the sign of

(x + 2)

depends on

x

Combine the terms over a common denominator.

\frac{(3 - x) - (x + 2)}{x + 2} \geq 0 \implies \frac{1 - 2x}{x + 2} \geq 0

Simplifying the numerator yields a clear rational inequality expression.

Identify the critical points and domain restrictions.

Numerator critical point:

x = \frac{1}{2}

; Denominator restriction:

x \neq -2

The quotient changes sign around

x = \frac{1}{2}

and

x = -2

, and division by zero is undefined.

Test the intervals

(-\infty, -2)

(-2, \frac{1}{2}]

, and

(\frac{1}{2}, \infty)

For

x \in (-2, \frac{1}{2}]

, the expression

\frac{1 - 2x}{x + 2}

is non-negative.

When

x = 0

\frac{1}{2} > 0

(positive). Outside this interval, the ratio is negative.

Solving Rational and Linear/Quadratic Inequalities

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