What is (3x+1)(2x-1)<= 3+x ?

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

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

(3 x + 1) (2 x - 1) \leq 3 + x

- \frac{2}{3} \leq x \leq 1

Interval Notation

[- \frac{2}{3}, 1]

Decimal Notation

- 0.66666 \dots \leq x \leq 1

Solution steps

(3 x + 1) (2 x - 1) \leq 3 + x

Rewrite in standard form

3 x^{2} - x - 2 \leq 0

Factor

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

(3 x + 2) (x - 1) \leq 0

Identify the intervals

- \frac{2}{3} \leq x \leq 1

2 (5 - x) + 4 < 3 x - 6 - 2 (x + 7)

\frac{10}{( - \frac{6}{11} )}

\frac{2}{1 - \frac{1}{5}}

2 x^{2} - 20 x + 100 = 0

- x^{2} + 36 = 0

What is (3x+1)(2x-1)<= 3+x ?
The solution to (3x+1)(2x-1)<= 3+x is -2/3 <= x<= 1