5
$\begingroup$

I keep trying to solve it but I seem to be making an error somewhere: $$4-\left|\frac{5y}{3}+4\right| > \frac25$$

  • 7
    The big hint is that $|a|2010-12-22

1 Answers 1

4

$4-\left|\frac{5y}{3}+4\right| > \frac25$ can be rewritten as $-\left|\frac{5y}{3}+4\right| > \frac{-18}{5}$ or $\left|\frac{5y}{3}+4\right| < \frac{18}{5}$. From here, represent the expression as $\frac{-18}{5} < \frac{5y}{3}+4\ < \frac{18}{5}$ and proceed to solve it using arithmetic.

  • 1
    you accidentally left the absolute value there in the last line2010-12-22