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