What is ((5x-1))/((x^2+3))<1 ?

The solution to ((5x-1))/((x^2+3)) 4

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((5x-1))/((x^2+3))<1

\frac{( 5 x - 1 )}{( x ^{2} + 3 )} < 1

x < 1 or x > 4

Interval Notation

(- \infty, 1) \cup (4, \infty)

Solution steps

\frac{5 x - 1}{x ^{2} + 3} < 1

Rewrite in standard form

\frac{- x ^{2} + 5 x - 4}{x ^{2} + 3} < 0

Factor

\frac{- x ^{2} + 5 x - 4}{x ^{2} + 3} : \frac{- ( x - 1 ) ( x - 4 )}{x ^{2} + 3}

\frac{- ( x - 1 ) ( x - 4 )}{x ^{2} + 3} < 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x - 1 ) ( x - 4 ) ) ( - 1 )}{x ^{2} + 3} > 0 \cdot (- 1)

Simplify

\frac{( x - 1 ) ( x - 4 )}{x ^{2} + 3} > 0

Identify the intervals

x < 1 or x > 4

9 x^{2} - 8 x - 1 > 0

\frac{1}{3 ( 9 x + 27 )} > x + 33

\frac{\frac{x}{l o g _{10} ( 1 )}}{2} (2 x + 3) > 0

\frac{( 4 x + 3 )}{( 2 x + 1 )} < 2

(x^{2} + 3 x + 1) (x^{2} + 3 x - 3) \geq 5

What is ((5x-1))/((x^2+3))<1 ?
The solution to ((5x-1))/((x^2+3))<1 is x<1\lor x>4