I keep trying to solve it but I seem to be making an error somewhere: $$4-\left|\frac{5y}{3}+4\right| > \frac25$$
what is the process to solving this absolute value inequality: $4-\left|\frac{5y}{3}+4\right| > \frac25$?
5
$\begingroup$
algebra-precalculus
inequality
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7The big hint is that $|a| – 2010-12-22
1 Answers
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$4-\left|\frac{5y}{3}+4\right| > \frac25$ 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.
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1you accidentally left the absolute value there in the last line – 2010-12-22