((x+3))/((x+16))>= 3/((x-4))

Lösung

\frac{( x + 3 )}{( x + 16 )} \geq \frac{3}{( x - 4 )}

x < - 16 or - 6 \leq x < 4 or x \geq 10

Intervall-Notation

(- \infty, - 16) \cup [- 6, 4) \cup [10, \infty)

Schritte zur Lösung

\frac{x + 3}{x + 16} \geq \frac{3}{x - 4}

Rewrite in standard form

\frac{x ^{2} - 4 x - 60}{( x - 4 ) ( x + 16 )} \geq 0

Faktorisiere

\frac{x ^{2} - 4 x - 60}{( x - 4 ) ( x + 16 )} : \frac{( x + 6 ) ( x - 10 )}{( x - 4 ) ( x + 16 )}

\frac{( x + 6 ) ( x - 10 )}{( x - 4 ) ( x + 16 )} \geq 0

Identifiziere die Intervalle

x < - 16 or - 6 \leq x < 4 or x \geq 10

x^{3} + 8 x^{2} < 0

\frac{( ( x - 1 ) ( x ^{2} - x + 12 ) )}{( 3 + x ^{2} )} \leq 0

\frac{( 2 x + 3 )}{( x - 3 )} \leq 4

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

\frac{1}{x ^{2}} < 4