Question
Solve the following linear in-equation and graph the solution set on a real number line:
$\frac{1}{3}(2 x-1)<\frac{1}{4}(x+5)<\frac{1}{6}(3 x+4), x \in R$
$\frac{1}{3}(2 x-1)<\frac{1}{4}(x+5)<\frac{1}{6}(3 x+4), x \in R$
✓
Answer
$\frac{1}{3}(2 x-1)<\frac{1}{4}(x+5)$
$4(2 x-1)<3(x+5)$
$8 x-4<3 x+15$
$8 x-3 x<15+4$
$5 x<19$
$x<3 \frac{4}{5}$
$\text { and }$
$\frac{1}{4}(x+5)<\frac{1}{6}(3 x+4)$
$6(x+5)<4(3 x+4)$
$6 x+30<12 x+16$
$6 x-12 x<16-30$
$-6 x<-14$
$x>2 \frac{1}{3} $
Solution set $=\left[2 \frac{1}{3}<x<3 \frac{4}{5}\right]$
$4(2 x-1)<3(x+5)$
$8 x-4<3 x+15$
$8 x-3 x<15+4$
$5 x<19$
$x<3 \frac{4}{5}$
$\text { and }$
$\frac{1}{4}(x+5)<\frac{1}{6}(3 x+4)$
$6(x+5)<4(3 x+4)$
$6 x+30<12 x+16$
$6 x-12 x<16-30$
$-6 x<-14$
$x>2 \frac{1}{3} $
Solution set $=\left[2 \frac{1}{3}<x<3 \frac{4}{5}\right]$
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