What is (x-5)/3+(x+4)/2 >= (x+3)/6 ?

The solution to (x-5)/3+(x+4)/2 >= (x+3)/6 is x>= 1/4

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(x-5)/3+(x+4)/2 >= (x+3)/6

\frac{x - 5}{3} + \frac{x + 4}{2} \geq \frac{x + 3}{6}

x \geq \frac{1}{4}

Interval Notation

[\frac{1}{4}, \infty)

Decimal Notation

x \geq 0.25

Solution steps

\frac{x - 5}{3} + \frac{x + 4}{2} \geq \frac{x + 3}{6}

Multiply by LCM

2 (x - 5) + 3 (x + 4) \geq x + 3

Expand

2 (x - 5) + 3 (x + 4) : 5 x + 2

5 x + 2 \geq x + 3

Move

2

to the right side

5 x \geq x + 1

Move

x

to the left side

4 x \geq 1

Divide both sides by

4

x \geq \frac{1}{4}

(x^{4} y^{2} z) (y^{2} z)

(x + 4 a)^{2} - 25 a^{2}

m^{2} + 21 m - 22 = 0

lo g_{4} (2) + lo g_{4} (32)

5 (y) (- y) (y^{2}) (- y^{2})

What is (x-5)/3+(x+4)/2 >= (x+3)/6 ?
The solution to (x-5)/3+(x+4)/2 >= (x+3)/6 is x>= 1/4