What is the answer to 1/(y-2)=(2y+1)/(y^2-4) ?

The answer to 1/(y-2)=(2y+1)/(y^2-4) is y=1

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1/(y-2)=(2y+1)/(y^2-4)

\frac{1}{y - 2} = \frac{2 y + 1}{y ^{2} - 4}

y = 1

Solution steps

\frac{1}{y - 2} = \frac{2 y + 1}{y ^{2} - 4}

Cross multiply

y^{2} - 4 = (y - 2) (2 y + 1)

Solve

y^{2} - 4 = (y - 2) (2 y + 1) : y = 2, y = 1

y = 2, y = 1

Verify Solutions

Find undefined (singularity) points:

y = 2, y = - 2

Since the equation is undefined for:

2

y = 1

\frac{15 - 2 x - x ^{2}}{x - 4} = 0

\frac{3 z - 6}{7 - 2 z} = \frac{1.2}{3.2}

\frac{x}{x - 6} - 8 = - \frac{1}{x - 6}

\frac{8}{x ^{2}} + 1 = \frac{9}{x}

\frac{1}{x + 9} - 1 = \frac{x}{x + 9}

What is the answer to 1/(y-2)=(2y+1)/(y^2-4) ?
The answer to 1/(y-2)=(2y+1)/(y^2-4) is y=1
Solve for y: 1/(y-2)=(2y+1)/(y^2-4)
The solution is y=1