What is 6+1/(1-x)<= 7/x ?

The solution to 6+1/(1-x)<= 7/x is 0<x<= (7-sqrt(7))/6 \lor 1<x<= (7+sqrt(7))/6

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6+1/(1-x)<= 7/x

6 + \frac{1}{1 - x} \leq \frac{7}{x}

0 < x \leq \frac{7 - 7}{6} or 1 < x \leq \frac{7 + 7}{6}

Interval Notation

(0, \frac{7 - 7}{6}] \cup (1, \frac{7 + 7}{6}]

Decimal Notation

0 < x \leq 0.72570 \dots or 1 < x \leq 1.60762 \dots

Solution steps

6 + \frac{1}{1 - x} \leq \frac{7}{x}

Rewrite in standard form

\frac{- 6 x ^{2} + 14 x - 7}{x ( - x + 1 )} \leq 0

Factor

- 6 x^{2} + 14 x - 7 : - (x - \frac{7 + 7}{6}) (x - \frac{7 - 7}{6})

\frac{- ( x - \frac{7 + 7}{6} ) ( x - \frac{7 - 7}{6} )}{x ( - x + 1 )} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x - \frac{7 + 7}{6} ) ( x - \frac{7 - 7}{6} ) ) ( - 1 )}{x ( - x + 1 )} \geq 0 \cdot (- 1)

Simplify

\frac{( x - \frac{7 + 7}{6} ) ( x - \frac{7 - 7}{6} )}{x ( - x + 1 )} \geq 0

Identify the intervals

0 < x \leq \frac{7 - 7}{6} or 1 < x \leq \frac{7 + 7}{6}

\frac{5}{4} m = 20

x^{2} - x y - 2 y^{2}

16 t^{7} + 250 t^{4}

7 + 3 x > 2 x - 5

x \cdot 3 = 4

What is 6+1/(1-x)<= 7/x ?
The solution to 6+1/(1-x)<= 7/x is 0<x<= (7-sqrt(7))/6 \lor 1<x<= (7+sqrt(7))/6