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Popular >

5|2x+1|-3>= 9

Solution

5 ∣ 2 x + 1 ∣ - 3 \geq 9

Solution

x \leq - \frac{17}{10} or x \geq \frac{7}{10}

+2

Interval Notation

(- \infty, - \frac{17}{10}] \cup [\frac{7}{10}, \infty)

Decimal Notation

x \leq - 1.7 or x \geq 0.7

Solution steps

5 ∣ 2 x + 1 ∣ - 3 \geq 9

Move

3

to the right side

5 ∣ 2 x + 1 ∣ \geq 12

Divide both sides by

5

∣ 2 x + 1 ∣ \geq \frac{12}{5}

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

or

u \geq a

2 x + 1 \leq - \frac{12}{5} or 2 x + 1 \geq \frac{12}{5}

2 x + 1 \leq - \frac{12}{5} or 2 x + 1 \geq \frac{12}{5}

x \leq - \frac{17}{10} or x \geq \frac{7}{10}

Popular Examples

- x^{2} - 2 x + 35 = 0

factor 8a^2+12a factor

8 a^{2} + 12 a

simplify 8(3)simplify

8 (3)

factor x^{12}-y^{12}factor

x^{12} - y^{12}

(6n)/5+1/2 =(4n-3)/(10)

\frac{6 n}{5} + \frac{1}{2} = \frac{4 n - 3}{10}

Frequently Asked Questions (FAQ)

What is 5|2x+1|-3>= 9 ?
The solution to 5|2x+1|-3>= 9 is x<=-17/10 \lor x>= 7/10