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{2}{x - 3} < 0

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

\frac{( x + 5 )}{( x ^{2} - 3 x - 4 )} > 0

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

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

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