Question

) \geqslant 15x - 20" alt=" - 3(9x + 20) \geqslant 15x - 20" align="absmiddle" class="latex-formula">
could you please help me?​

Answer 1

Answer:

$x\le \:-\frac{20}{21}$

Step-by-step explanation:

$-3\left(9x+20\right)\ge \:15x-20\\\\\mathrm{Expand\:}-3\left(9x+20\right):\quad -27x-60\\-27x-60\ge \:15x-20\\\\\mathrm{Add\:}60\mathrm{\:to\:both\:sides}\\-27x-60+60\ge \:15x-20+60\\\\\mathrm{Simplify}\\-27x\ge \:15x+40\\\\\mathrm{Subtract\:}15x\mathrm{\:from\:both\:sides}\\-27x-15x\ge \:15x+40-15x\\\\\mathrm{Simplify}\\-42x\ge \:40\\\\\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}\\\left(-42x\right)\left(-1\right)\le \:40\left(-1\right)$

$Simplify\\42x\le \:-40\\\\\mathrm{Divide\:both\:sides\:by\:}42\\\frac{42x}{42}\le \frac{-40}{42}\\\\Simplify\\x\le \:-\frac{20}{21}$