What is (|1-3x|)/(x^2+1)>= 1 ?

The solution to (|1-3x|)/(x^2+1)>= 1 is True for all x

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(|1-3x|)/(x^2+1)>= 1

\frac{∣ 1 - 3 x ∣}{x ^{2} + 1} \geq 1

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

\frac{∣ 1 - 3 x ∣}{x ^{2} + 1} \geq 1

Move

1

to the left side

\frac{∣ 1 - 3 x ∣}{x ^{2} + 1} - 1 \geq 0

∣ 1 - 3 x ∣ \geq 0

\frac{a}{b} \geq 0 and a \geq 0

then

b > 0 or a = 0

x^{2} + 1 > 0 or ∣ 1 - 3 x ∣ = 0

x^{2} + 1 > 0 :

True for all

x

∣ 1 - 3 x ∣ = 0 : x = \frac{1}{3}

Combine the intervals

True for all x or x = \frac{1}{3}

Merge Overlapping Intervals

True for all x

1^{2} + 1^{2}

- 5 a + 12 < - 18

200 x - 2 x^{2} = - x^{2} + 10 x + 4125

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

5 x^{2} + 5 x + 8 = 0

What is (|1-3x|)/(x^2+1)>= 1 ?
The solution to (|1-3x|)/(x^2+1)>= 1 is True for all x