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

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

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((x+3))/((x+1))<5

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

x < - 1 or x > - \frac{1}{2}

Interval Notation

(- \infty, - 1) \cup (- \frac{1}{2}, \infty)

Decimal Notation

x < - 1 or x > - 0.5

Solution steps

\frac{x + 3}{x + 1} < 5

Rewrite in standard form

\frac{- 4 x - 2}{x + 1} < 0

Factor

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

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Divide both sides by

2

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

Simplify

\frac{2 x + 1}{x + 1} > 0

Identify the intervals

x < - 1 or x > - \frac{1}{2}

x \cdot (x + 2)^{2} \cdot (x - 1)^{5} \cdot (2 x - 3) \cdot (x - 3)^{4} \geq 0

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

\frac{( ( 2 x + 1 ) ( x + 4 ) )}{( x ^{2} )} \geq 0

(x^{2} - 16) \cdot (x^{2} - 4) < 0

(x - 2) (x - 2) - 15 < x - 2 + x - 2

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