What is (x-3)/(x^2+9)<= x/(x^2+3) ?

The solution to (x-3)/(x^2+9)<= x/(x^2+3) is True for all x

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(x-3)/(x^2+9)<= x/(x^2+3)

\frac{x - 3}{x ^{2} + 9} \leq \frac{x}{x ^{2} + 3}

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

\frac{x - 3}{x ^{2} + 9} \leq \frac{x}{x ^{2} + 3}

Rewrite in standard form

\frac{- 3 x ^{2} - 6 x - 9}{( x ^{2} + 9 ) ( x ^{2} + 3 )} \leq 0

Factor

\frac{- 3 x ^{2} - 6 x - 9}{( x ^{2} + 9 ) ( x ^{2} + 3 )} : \frac{- 3 ( x ^{2} + 2 x + 3 )}{( x ^{2} + 9 ) ( x ^{2} + 3 )}

\frac{- 3 ( x ^{2} + 2 x + 3 )}{( x ^{2} + 9 ) ( x ^{2} + 3 )} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - 3 ( x ^{2} + 2 x + 3 ) ) ( - 1 )}{( x ^{2} + 9 ) ( x ^{2} + 3 )} \geq 0 \cdot (- 1)

Simplify

\frac{3 ( x ^{2} + 2 x + 3 )}{( x ^{2} + 9 ) ( x ^{2} + 3 )} \geq 0

Divide both sides by

3

\frac{\frac{3 ( x ^{2} + 2 x + 3 )}{( x ^{2} + 9 ) ( x ^{2} + 3 )}}{3} \geq \frac{0}{3}

Simplify

\frac{x ^{2} + 2 x + 3}{( x ^{2} + 9 ) ( x ^{2} + 3 )} \geq 0

Identify the intervals

True for all x

7 x^{2} - 6 = - 35

∣ 3 x - 6 ∣ \geq 4

2 m^{2} - 7 m - 660 = 0

6 x - 2 \leq 3 x - 6

4 (x + 2) = - 2 (x - 1)

What is (x-3)/(x^2+9)<= x/(x^2+3) ?
The solution to (x-3)/(x^2+9)<= x/(x^2+3) is True for all x