What is (7x}{x+1}<= \frac{6x+3)/4 ?

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(7x}{x+1}<= \frac{6x+3)/4

\frac{7 x}{x + 1} \leq \frac{6 x + 3}{4}

- 1 < x \leq \frac{1}{6} or x \geq 3

Interval Notation

(- 1, \frac{1}{6}] \cup [3, \infty)

Decimal Notation

- 1 < x \leq 0.16666 \dots or x \geq 3

Solution steps

\frac{7 x}{x + 1} \leq \frac{6 x + 3}{4}

Rewrite in standard form

\frac{- 6 x ^{2} + 19 x - 3}{x + 1} \leq 0

Factor

\frac{- 6 x ^{2} + 19 x - 3}{x + 1} : \frac{- ( 6 x - 1 ) ( x - 3 )}{x + 1}

\frac{- ( 6 x - 1 ) ( x - 3 )}{x + 1} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( 6 x - 1 ) ( x - 3 ) ) ( - 1 )}{x + 1} \geq 0 \cdot (- 1)

Simplify

\frac{( 6 x - 1 ) ( x - 3 )}{x + 1} \geq 0

Identify the intervals

- 1 < x \leq \frac{1}{6} or x \geq 3

- 8 n - 13 \leq 27

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

\frac{( x ^{2} - 27 )}{( a ^{3} - 1 )}

\frac{2 a}{9 - a ^{2}} \cdot \frac{a - 3}{a ^{3} + a}

6 a - 3 > 21

What is (7x}{x+1}<= \frac{6x+3)/4 ?
The solution to (7x}{x+1}<= \frac{6x+3)/4 is -1<x<= 1/6 \lor x>= 3