What is (x+1)/(2x-4)< 1/(1/3 x-1) ?

The solution to (x+1)/(2x-4)< 1/(1/3 x-1) is 4-sqrt(7)<x<2\lor 3<x<4+sqrt(7)

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(x+1)/(2x-4)< 1/(1/3 x-1)

\frac{x + 1}{2 x - 4} < \frac{1}{\frac{1}{3} x - 1}

4 - 7 < x < 2 or 3 < x < 4 + 7

Interval Notation

(4 - 7, 2) \cup (3, 4 + 7)

Decimal Notation

1.35424 \dots < x < 2 or 3 < x < 6.64575 \dots

Solution steps

\frac{x + 1}{2 x - 4} < \frac{1}{\frac{1}{3} x - 1}

Rewrite in standard form

\frac{x ^{2} - 8 x + 9}{( x - 2 ) ( x - 3 )} < 0

Factor

x^{2} - 8 x + 9 : (x - 4 - 7) (x - 4 + 7)

\frac{( x - 4 - 7 ) ( x - 4 + 7 )}{( x - 2 ) ( x - 3 )} < 0

Identify the intervals

4 - 7 < x < 2 or 3 < x < 4 + 7

\frac{( 8 k ^{- 5} ) ( 2 k ^{3} )}{4 k ^{- 6}}

\frac{3}{- 3}

(\frac{1}{9})^{\frac{1}{2}}

4 x^{2} - 4 x y + y^{2}

a^{3} b^{6} - b^{3}

What is (x+1)/(2x-4)< 1/(1/3 x-1) ?
The solution to (x+1)/(2x-4)< 1/(1/3 x-1) is 4-sqrt(7)<x<2\lor 3<x<4+sqrt(7)