What is the answer to solvefor x,f(x-3)=((x^2-9))/((x+1)) ?

The answer to solvefor x,f(x-3)=((x^2-9))/((x+1)) is x=3,x=(-f+3)/(f-1);\quad f\ne 1

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solvefor x,f(x-3)=((x^2-9))/((x+1))

solve for

x, f (x - 3) = \frac{( x ^{2} - 9 )}{( x + 1 )}

x = 3, x = \frac{- f + 3}{f - 1}; f \neq = 1

Solution steps

f (x - 3) = \frac{( x ^{2} - 9 )}{( x + 1 )}

Simplify

\frac{( x ^{2} - 9 )}{( x + 1 )} : \frac{x ^{2} - 9}{x + 1}

f (x - 3) = \frac{x ^{2} - 9}{x + 1}

Multiply both sides by

x + 1

f (x - 3) (x + 1) = x^{2} - 9

Solve

f (x - 3) (x + 1) = x^{2} - 9 : x = 3, x = \frac{- f + 3}{f - 1}; f \neq = 1

x = 3, x = \frac{- f + 3}{f - 1}; f \neq = 1

Verify Solutions

Find undefined (singularity) points:

x = - 1

Combine undefined points with solutions:

x = 3, x = \frac{- f + 3}{f - 1}; f \neq = 1

80 x^{- 2} = 6 x^{- 1} + 2

(\frac{1 - 1}{x})^{10} = 0.8

x, \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1

\frac{( x + 1 )}{( 3 x - 1 )} = (\frac{20}{50})

x^{2} + [\frac{x}{( x + 1 )}]^{2} = 3

What is the answer to solvefor x,f(x-3)=((x^2-9))/((x+1)) ?
The answer to solvefor x,f(x-3)=((x^2-9))/((x+1)) is x=3,x=(-f+3)/(f-1);\quad f\ne 1