((x-3))/5 >= ((3-x))/((8x))

Lösung

\frac{( x - 3 )}{5} \geq \frac{( 3 - x )}{( 8 x )}

- \frac{5}{8} \leq x < 0 or x \geq 3

Intervall-Notation

[- \frac{5}{8}, 0) \cup [3, \infty)

Dezimale

- 0.625 \leq x < 0 or x \geq 3

Schritte zur Lösung

\frac{x - 3}{5} \geq \frac{3 - x}{8 x}

Rewrite in standard form

\frac{8 x ^{2} - 19 x - 15}{x} \geq 0

Faktorisiere

\frac{8 x ^{2} - 19 x - 15}{x} : \frac{( 8 x + 5 ) ( x - 3 )}{x}

\frac{( 8 x + 5 ) ( x - 3 )}{x} \geq 0

Identifiziere die Intervalle

- \frac{5}{8} \leq x < 0 or x \geq 3

\frac{( x ^{2} - 5 x + 6 )}{( x ^{2} + x + 1 )} < 0

\frac{( ( x - 1 ) ( x - 2 ) )}{( x + 5 )} > 1

\frac{( x - 2 )}{( x + 2 )} \geq 3, 2 x - 7 \leq 5

\frac{( x ^{3} - x ^{2} + x - 1 )}{( x + 8 )} \leq 0

\frac{( 2 x - 4 )}{x} \geq 7