Question

What are the solutions of the inequality? -2(3x + 2) ≥ -6x - 4 Question 2 options: all real numbers x ≤ 6 no solution x ≥ 0

Answer 1

$Solution, -2\left(3x+2\right)\ge \:-6x-4\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:True\quad \forall \:x\in \mathbb{R}\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}$

$Steps:$

$-2\left(3x+2\right)\ge \:-6x-4$

$\mathrm{Expand\:}-2\left(3x+2\right),\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac,\\a=-2,\:b=3x,\:c=2,\\-2\cdot \:3x+\left(-2\right)\cdot \:2,\\\mathrm{Apply\:minus-plus\:rules},\\+\left(-a\right)=-a,\\-2\cdot \:3x-2\cdot \:2,\\\mathrm{Simplify}\:-2\cdot \:3x-2\cdot \:2,\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:3=6,\\-6x-2\cdot \:2,\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4,\\-6x-4,\\-6x-4\ge \:-6x-4$

$\mathrm{Add\:}4\mathrm{\:to\:both\:sides},\\-6x-4+4\ge \:-6x-4+4$

$\mathrm{Simplify},\\-6x\ge \:-6x$

$\mathrm{Add\:}6x\mathrm{\:to\:both\:sides},\\-6x+6x\ge \:-6x+6x$

$\mathrm{Simplify},\\0\ge \:0$

$\mathrm{Therefore,\:the\:final\:solution\:is},\\True\quad \forall \:x\in \mathbb{R}$

$Or\:No\:Solution\:x\geq 0$