Question

If $\frac{2x-10}{x+2}\ge x-5$ then $x$ lies in

• A. $(-\infty,-1)\cup[0,6]$
• B. $(-\infty,-2)\cup[-1,5]$
• C. $(-\infty,-2)\cup[0,10]$
• D. $(-\infty,0)\cup[0,5]$
• E. $(-\infty,-2)\cup[0,5]$

Hint:

Math Tip: Never cross-multiply variables in an inequality! Doing so assumes the denominator is positive, which can eliminate valid solutions or introduce invalid ones. Always bring everything to one side and use the Wavy Curve Method.

Answer 1

Answer: (E)

Concept:
Algebra - Solving Rational Inequalities (Wavy Curve Method).
Step 1: Bring all terms to one side of the inequality.
Given inequality: $$ \frac{2x-10}{x+2} \ge x-5 $$ Subtract $(x-5)$ from both sides: $$ \frac{2x-10}{x+2} - (x-5) \ge 0 $$
Step 2: Factor the numerator and simplify.
Notice that $2x-10$ can be factored as $2(x-5)$: $$ \frac{2(x-5)}{x+2} - \frac{(x-5)(x+2)}{x+2} \ge 0 $$ Factor out the common term $(x-5)$: $$ (x-5) \left[ \frac{2 - (x+2)}{x+2} \right] \ge 0 $$ $$ (x-5) \left[ \frac{-x}{x+2} \right] \ge 0 $$
Step 3: Rearrange to standard form.
Multiply the entire inequality by $-1$. Remember that multiplying by a negative number flips the inequality sign: $$ \frac{x(x-5)}{x+2} \le 0 $$
Step 4: Find the critical points.
Set the numerator and denominator factors to zero to find the critical points:

• Numerator: $x = 0$ and $x = 5$ (These can be included, so we use closed brackets).
• Denominator: $x+2 = 0 \Rightarrow x = -2$ (This cannot be included because division by zero is undefined, so we use an open parenthesis).

Step 5: Apply the Wavy Curve Method.
Plot the critical points $-2, 0, 5$ on a number line. Check the sign of the expression $f(x) = \frac{x(x-5)}{x+2}$ in each interval:

• For $x>5$ (e.g., $x=6$): $f(x) = \frac{(+)(+)}{+}>0$
• For $0<x<5$ (e.g., $x=1$): $f(x) = \frac{(+)(-)}{+}<0$
• For $-2<x<0$ (e.g., $x=-1$): $f(x) = \frac{(-)(-)}{+}>0$
• For $x<-2$ (e.g., $x=-3$): $f(x) = \frac{(-)(-)}{-}<0$

Step 6: Determine the final interval.
We need the intervals where $f(x) \le 0$ (negative or zero). Looking at the sign scheme, this occurs in the intervals $(-\infty, -2)$ and $[0, 5]$. The solution set is the union of these intervals: $$ x \in (-\infty,-2) \cup [0,5] $$