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Solve \(\displaystyle \frac{1}{y + 2} - \frac{4}{2y + 1} + \frac{3y - 4}{2y^2 + 5y + 2} = 0\)

 May 7, 2020
 #1
avatar+50 
+1

we note that \((y+2)(2y+1) = 2y^2 + 5y +2\), so we plug that in and solve for y:

\(\begin{align*} \frac{1}{y+2}-\frac{4}{2y+1} + \frac{3y-4}{2y^2 + 5y + 2}&= 0 \\ \frac{2y+1}{2y^2 + 5y + 2} - \frac{4(y+2)}{2y^2 + 5y + 2} + \frac{3y-4}{2y^2 + 5y + 2}&= 0 \\ \frac{2y+1-4y-8+3y-4}{2y^2 + 5y + 2}&=0 \\ \frac{y-11}{2y^2 + 5y + 2}&=0 \\ y-11&=0 \\ y&=\boxed{11} \end{align*} \)

note: since we multiply by \(2y^2 + 5y +2\), we have \(y \neq -2, -\frac{1}{2}\), but since \(y+2\) and \(2y+1\) are already denominators, we don't have to worry about fractions over zero.

 May 7, 2020
 #2
avatar+23687 
+1

 [ (2y+1)(y+2)][   1/(y+2)  - 4/(2y+1) + (3y-4)/(2y^2+5y+2)]  =

2y+1  -    4y-8   +    3y-4    = 0

y -11=0

y=11

 May 7, 2020

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