What is ((10x))/((x^2+9))<= 1 ?

The solution to ((10x))/((x^2+9)) = 9

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((10x))/((x^2+9))<= 1

\frac{( 10 x )}{( x ^{2} + 9 )} \leq 1

x \leq 1 or x \geq 9

Interval Notation

(- \infty, 1] \cup [9, \infty)

Solution steps

\frac{10 x}{x ^{2} + 9} \leq 1

Rewrite in standard form

\frac{10 x - x ^{2} - 9}{x ^{2} + 9} \leq 0

Factor

\frac{10 x - x ^{2} - 9}{x ^{2} + 9} : \frac{- ( x - 1 ) ( x - 9 )}{x ^{2} + 9}

\frac{- ( x - 1 ) ( x - 9 )}{x ^{2} + 9} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x - 1 ) ( x - 9 ) ) ( - 1 )}{x ^{2} + 9} \geq 0 \cdot (- 1)

Simplify

\frac{( x - 1 ) ( x - 9 )}{x ^{2} + 9} \geq 0

Identify the intervals

x \leq 1 or x \geq 9

\frac{e}{x} > 0

\frac{( 3 x ^{2} - 4 x - 6 )}{( 2 x - 5 )} < 0

\frac{( x ^{2} - 7 x + 12 )}{( 2 x ^{2} + 4 x + 5 )} > \frac{1}{2}

\frac{( x ^{4} )}{( ( x - 2 ) ^{2} )} > 0

\frac{( ( x - 1 ) ( x - 5 ) )}{( x - 3 )} > 0

What is ((10x))/((x^2+9))<= 1 ?
The solution to ((10x))/((x^2+9))<= 1 is x<= 1\lor x>= 9