What is (6x-3)/(x-1)>= x ?

The solution to (6x-3)/(x-1)>= x is x<= (7-sqrt(37))/2 \lor 1<x<= (7+sqrt(37))/2

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(6x-3)/(x-1)>= x

\frac{6 x - 3}{x - 1} \geq x

x \leq \frac{7 - 37}{2} or 1 < x \leq \frac{7 + 37}{2}

Interval Notation

(- \infty, \frac{7 - 37}{2}] \cup (1, \frac{7 + 37}{2}]

Decimal Notation

x \leq 0.45861 \dots or 1 < x \leq 6.54138 \dots

Solution steps

\frac{6 x - 3}{x - 1} \geq x

Rewrite in standard form

\frac{- x ^{2} + 7 x - 3}{x - 1} \geq 0

Factor

- x^{2} + 7 x - 3 : - (x - \frac{7 + 37}{2}) (x - \frac{7 - 37}{2})

\frac{- ( x - \frac{7 + 37}{2} ) ( x - \frac{7 - 37}{2} )}{x - 1} \geq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x - \frac{7 + 37}{2} ) ( x - \frac{7 - 37}{2} ) ) ( - 1 )}{x - 1} \leq 0 \cdot (- 1)

Simplify

\frac{( x - \frac{7 + 37}{2} ) ( x - \frac{7 - 37}{2} )}{x - 1} \leq 0

Identify the intervals

x \leq \frac{7 - 37}{2} or 1 < x \leq \frac{7 + 37}{2}

2 x^{2} - 5 x + 3 \geq 0

4 - k^{2}

t = 2 (9) + 9

x^{2} + 6 x - 112 = 0

h^{2} + 7 h = 44

What is (6x-3)/(x-1)>= x ?
The solution to (6x-3)/(x-1)>= x is x<= (7-sqrt(37))/2 \lor 1<x<= (7+sqrt(37))/2