Question

determine the range of values of x that satisfy the inequality

Answer 1

Step-by-step explanation:

Given that:

$\bigg | \frac{x + 3}{x}\bigg | \geqslant \bigg| \frac{x}{2 - x} \bigg|$

To find: Determine the range of values of x that satisfy the inequality

Solution:

To find range of values of x,first open

Absolute value

Case1:

$\frac{x + 3}{x} \geqslant \frac{x}{2 - x} \\ \\ (x + 3)(2 - x) \geqslant {x}^{2} \\ \\ 2x - {x}^{2} + 6 - 3x \geqslant {x}^{2} \\ \\ - 2 {x}^{2} - x + 6 \geqslant 0 \\ \\ - 2 {x}^{2} - 4x + 3x + 6 \geqslant 0 \\ \\ - 2x(x + 2) + 3(x + 2) \geqslant 0 \\ \\ (x + 2)( - 2x + 3) \geqslant 0 \\ \\ (x + 2) \geqslant 0 \\ \\ x \geqslant - 2 \\\\or\\ \\ - 2x + 3 \geqslant 0 \\ \\ - 2x \geqslant - 3 \\ \\ x \leqslant \frac{3}{2}$

Case2:

$\frac{x + 3}{x} \geqslant \frac{ - x}{2 - x} \\ \\ (x + 3)(2 - x) \geqslant - {x}^{2} \\ \\ - {x}^{2} + 2x + 6 - 3x \geqslant - {x}^{2} \\ \\ - x + 6 \geqslant 0 \\ \\ x \geqslant 6$

Thus,

range of values that satisfies the given inequality

$- 2 \leqslant x < 0 \: \: or \: \: 0 < x \leqslant \frac{3}{2} \: \: or \: x \geqslant 6$

In interval notation

$[- 2 ,\: 0) \cup(0, \: \frac{3}{2}] \cup[6 \:,∞ )$

Hope it helps you.