What is the solution to the inequality?
Express yourself in interval notation.
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Answer:
[tex]\begin{bmatrix}\mathrm{Solution:}\:&\:x\ge \frac{20}{23}\:\\ \:\mathrm{Decimal:}&\:x\ge \:0.86956\dots \\ \:\mathrm{Interval\:Notation:}&\:[\frac{20}{23},\:\infty \:)\end{bmatrix}[/tex]
Step-by-step explanation:
[tex]-\frac{2}{3}\left(2x-\frac{1}{2}\right)\le \frac{1}{5}x-1[/tex]
Expand [tex]-\frac{2}{3}\left(2x-\frac{1}{2}\right)\le \frac{1}{5}x-1[/tex] [tex]-\frac{4}{3}x+\frac{1}{3}[/tex]
[tex]-\frac{4}{3}x+\frac{1}{3}\le \frac{1}{5}x-1[/tex]
[tex]\mathrm{Subtract\:}\frac{1}{3}\mathrm{\:from\:both\:sides}[/tex]
[tex]-\frac{4}{3}x+\frac{1}{3}-\frac{1}{3}\le \frac{1}{5}x-1-\frac{1}{3}[/tex]
[tex]Simplify[/tex]
[tex]-\frac{4}{3}x\le \:-\frac{4}{3}+\frac{1}{5}x[/tex]
[tex]\mathrm{Subtract\:}\frac{1}{5}x\mathrm{\:from\:both\:sides}[/tex]
[tex]-\frac{4}{3}x-\frac{1}{5}x\le \:-\frac{4}{3}+\frac{1}{5}x-\frac{1}{5}x[/tex]
[tex]Simplify[/tex]
[tex]-\frac{23}{15}x\le \:-\frac{4}{3}[/tex]
[tex]\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}[/tex]
[tex]\left(-\frac{23}{15}x\right)\left(-1\right)\ge \left(-\frac{4}{3}\right)\left(-1\right)[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]\frac{23}{15}x\ge \frac{4}{3}[/tex]
[tex]\mathrm{Multiply\:both\:sides\:by\:}15[/tex]
[tex]15\cdot \frac{23}{15}x\ge \frac{4\cdot \:15}{3}[/tex]
[tex]Simplify[/tex]
[tex]23x\ge \:20[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}23[/tex]
[tex]\frac{23x}{23}\ge \frac{20}{23}[/tex]
[tex]\mathrm{Simplify}[/tex]
Hence the final answer is [tex]\begin{bmatrix}\mathrm{Solution:}\:&\:x\ge \frac{20}{23}\:\\ \:\mathrm{Decimal:}&\:x\ge \:0.86956\dots \\ \:\mathrm{Interval\:Notation:}&\:[\frac{20}{23},\:\infty \:)\end{bmatrix}[/tex]