5|x|+6>= 1

Solution

5 ∣ x ∣ + 6 \geq 1

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

5 ∣ x ∣ + 6 \geq 1

Move

6

to the right side

5 ∣ x ∣ \geq - 5

Divide both sides by

5

∣ x ∣ \geq - 1

Absolute values are always greater or equal to 0

True for all x

x^{2} - 10 x + 8 = 0

(a - b + 1)^{2} - (1 - a - b)^{2}

\frac{1}{\frac{π}{2} - ( - \frac{π}{2} )}

18 x^{2} - 9 x - 1 = 0

20 a^{4} b - 20 a^{3} b + 5 a^{2} b