What is (4x^2+2x-6)/(3x+1)<3x+1 ?

The solution to (4x^2+2x-6)/(3x+1) -1/3

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(4x^2+2x-6)/(3x+1)<3x+1

\frac{4 x ^{2} + 2 x - 6}{3 x + 1} < 3 x + 1

x > - \frac{1}{3}

Interval Notation

(- \frac{1}{3}, \infty)

Decimal Notation

x > - 0.33333 \dots

Solution steps

\frac{4 x ^{2} + 2 x - 6}{3 x + 1} < 3 x + 1

Rewrite in standard form

\frac{- 5 x ^{2} - 4 x - 7}{3 x + 1} < 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( 5 x ^{2} + 4 x + 7 ) ) ( - 1 )}{3 x + 1} > 0 \cdot (- 1)

Simplify

\frac{5 x ^{2} + 4 x + 7}{3 x + 1} > 0

Identify the intervals

x > - \frac{1}{3}

- 5 x^{2} + 7 x + 2 = 0

(3^{2})^{\frac{3}{2}}

\frac{3}{8}

20 x^{2} - 13 x - 15 = 0

\frac{w}{5} \leq 4

What is (4x^2+2x-6)/(3x+1)<3x+1 ?
The solution to (4x^2+2x-6)/(3x+1)<3x+1 is x>-1/3