Question

\(\frac{2x+\frac{3x-4}{5}}{15}\) < \(\frac{\frac{3-x}{2}+7x}{5}\) + 1 - x

Answer 1

* \(\frac{2x+\frac{3x-4}{5}}{15}=\frac{\frac{10x+3x-4}{5}}{15}=\frac{13x-4}{5}.\frac{1}{15}=\frac{13x-4}{75}\)

*\(\frac{\frac{3-x}{2}+7x}{5}=\frac{\frac{3-x+14x}{2}}{5}=\frac{\frac{3+13x}{2}}{5}=\left(\frac{3+13x}{2}.\frac{1}{5}\right)=\frac{3+13x}{10}\)

➝\(\frac{\frac{3-x}{2}+7x}{5}+1-x=\frac{3+13x}{10}+1-x=\frac{3+13x+10\left(1-x\right)}{10}=\frac{13+3x}{10}\)BPT⇔\(\frac{13x-4}{75}< \frac{13+3x}{10}\)

⇔\(\frac{13x}{75}+\frac{3x}{10}< \frac{13}{10}+\frac{4}{75}\)

⇔\(\frac{-19x}{150}< \frac{203}{150}\)

⇔\(x>\frac{-203}{19}\approx-10.68\)