Question

$$\frac{(0.4y^{-3})}{(1.5y+9)}=\frac{-7}{5}$$

Answer

y=-5.9991172790527,-0.32252426147461

Solution


Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[\frac{0.4\times \frac{1}{{y}^{3}}}{1.5y+9}=-\frac{7}{5}\]
Simplify  \(0.4\times \frac{1}{{y}^{3}}\)  to  \(\frac{0.4}{{y}^{3}}\).
\[\frac{\frac{0.4}{{y}^{3}}}{1.5y+9}=-\frac{7}{5}\]
Simplify  \(\frac{\frac{0.4}{{y}^{3}}}{1.5y+9}\)  to  \(\frac{0.4}{{y}^{3}(1.5y+9)}\).
\[\frac{0.4}{{y}^{3}(1.5y+9)}=-\frac{7}{5}\]
Multiply both sides by \({y}^{3}(1.5y+9)\).
\[0.4=-\frac{7}{5}{y}^{3}(1.5y+9)\]
Simplify  \(\frac{7}{5}{y}^{3}(1.5y+9)\)  to  \(\frac{7{y}^{3}(1.5y+9)}{5}\).
\[0.4=-\frac{7{y}^{3}(1.5y+9)}{5}\]
Multiply both sides by \(5\).
\[2=-7{y}^{3}(1.5y+9)\]
Expand.
\[2=-10.5{y}^{4}-63{y}^{3}\]
Move all terms to one side.
\[2+10.5{y}^{4}+63{y}^{3}=0\]
No root was found algebraically. However, the following root(s) were found by numerical methods.
\[y=-5.999117,-0.322524\]