Question

How do you solve $\dfrac{{{y^2} + 5y - 6}}{{{y^3} - 2{y^2}}} = \dfrac{5}{y} - \dfrac{6}{{{y^3} - 2{y^2}}}$ ?

Answer 1

Hint: To solve the given equation, to eliminate the complicated terms, we will multiply both sides by a factor that is able to eliminate the denominator. And then we will solve until we will solve for the variable $y$.

Complete step by step answer:
Given- $\dfrac{{{y^2} + 5y - 6}}{{{y^3} - 2{y^2}}} = \dfrac{5}{y} - \dfrac{6}{{{y^3} - 2{y^2}}}$
Restrict the domain to prevent division by 0.
$\dfrac{{{y^2} + 5y - 6}}{{{y^3} - 2{y^2}}} = \dfrac{5}{y} - \dfrac{6}{{{y^3} - 2{y^2}}}$ ; $y \ne 0,y \ne 2$
Multiply both sides by a factor that eliminates the denominators:
$\dfrac{{{y^2} + 5y - 6}}{{{y^3} - 2{y^2}}}({y^3} - 2{y^2}) = \dfrac{5}{y}({y^3} - 2{y^2}) - \dfrac{6}{{{y^3} - 2{y^2}}}({y^3} - 2{y^2})$$ \Rightarrow \dfrac{{{y^2} + 5y - 6}}{{{y^3} - 2{y^2}}}({y^3} - 2{y^2}) = \dfrac{5}{y}(y({y^3} - 2{y^2})) - \dfrac{6}{{{y^3} - 2{y^2}}}({y^3} - 2{y^2})$
Now cancel out the possible divisions in both numerator and denominator:
$\begin{align}
   \Rightarrow {y^2} + 5y - 6 = 5({y^2} - 2y) - 6\,;\,\,y \ne 0,y \ne 2 \\
   \Rightarrow {y^2} + 5y - 6 = 5{y^2} - 10y - 6\,;\,y \ne 0,y \ne 2 \\
   \Rightarrow 5{y^2} - 10y - 6 - {y^2} - 5y + 6 = 0\,;\,y \ne 0,y \ne 2 \\
   \Rightarrow 4{y^2} - 15y = 0\,;\,y \ne 0,y \ne 2 \\
\end{align} $
Solving the resulting quadratic equation:
$\therefore 4{y^2} - 15y = 0\,;\,y \ne 0,y \ne 2$
Now, we will take $y$ as common-
$y(4y - 15) = 0$
$\begin{align}
  \because 4y - 15 = 0 \\
   \Rightarrow y = \dfrac{{15}}{4} \\
\end{align} $

Note: During solving for $y$ ; as we can’t solve $y = 0$ ; so we discard the $y$ factor. A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real. It may be possible to express a quadratic equation \[a{x^2} + bx + c = 0\] as a product \[\left( {px{\text{ }} + {\text{ }}q} \right)\left( {rx{\text{ }} + {\text{ }}s} \right) = 0\] .