What is (2x^2)/3-x<(8x)/3 (1+x)+1 ?

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

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(2x^2)/3-x<(8x)/3 (1+x)+1

\frac{2 x ^{2}}{3} - x < \frac{8 x}{3} (1 + x) + 1

x < - \frac{3}{2} or x > - \frac{1}{3}

Solution steps

\frac{2 x ^{2}}{3} - x < \frac{8 x}{3} (1 + x) + 1

Rewrite in standard form

- 6 x^{2} - 11 x - 3 < 0

Factor

- 6 x^{2} - 11 x - 3 : - (3 x + 1) (2 x + 3)

- (3 x + 1) (2 x + 3) < 0

Multiply both sides by

- 1

(reverse the inequality)

(- (3 x + 1) (2 x + 3)) (- 1) > 0 \cdot (- 1)

Simplify

(3 x + 1) (2 x + 3) > 0

Identify the intervals

x < - \frac{3}{2} or x > - \frac{1}{3}

3 r - 6 = 18

8 1^{\frac{5}{2}}

1.6 < 1.9 y + 5.4

3 x^{2} - 6 x = 15

3 - 2 i

What is (2x^2)/3-x<(8x)/3 (1+x)+1 ?
The solution to (2x^2)/3-x<(8x)/3 (1+x)+1 is x<-3/2 \lor x>-1/3