What is the answer to (y^2-5}{3y}=\frac{-1)/3 ?

The answer to (y^2-5}{3y}=\frac{-1)/3 is y=(-1+sqrt(21))/2 ,y=-(1+sqrt(21))/2

Solve for y: (y^2-5}{3y}=\frac{-1)/3

The solution is y=(-1+sqrt(21))/2 ,y=-(1+sqrt(21))/2

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\frac{y ^{2} - 5}{3 y} = \frac{- 1}{3}

y = \frac{- 1 + 21}{2}, y = - \frac{1 + 21}{2}

Decimal Notation

y = 1.79128 \dots, y = - 2.79128 \dots

Solution steps

\frac{y ^{2} - 5}{3 y} = \frac{- 1}{3}

Cross multiply

(y^{2} - 5) \cdot 3 = - 3 y

Solve

(y^{2} - 5) \cdot 3 = - 3 y : y = \frac{- 1 + 21}{2}, y = - \frac{1 + 21}{2}

y = \frac{- 1 + 21}{2}, y = - \frac{1 + 21}{2}

Verify Solutions

Find undefined (singularity) points:

y = 0

Combine undefined points with solutions:

y = \frac{- 1 + 21}{2}, y = - \frac{1 + 21}{2}

\frac{13 y ^{2} + 12}{y ^{2}} = 44

\frac{3}{( x + 2 )} (x - 2) + \frac{2}{x} + 2 = \frac{1}{x} - 2

\frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{91}{37}

(\frac{1}{3 \cdot 2 x})^{4} = \frac{1}{16}

\frac{4}{y} = 2

What is the answer to (y^2-5}{3y}=\frac{-1)/3 ?
The answer to (y^2-5}{3y}=\frac{-1)/3 is y=(-1+sqrt(21))/2 ,y=-(1+sqrt(21))/2
Solve for y: (y^2-5}{3y}=\frac{-1)/3
The solution is y=(-1+sqrt(21))/2 ,y=-(1+sqrt(21))/2