What is x/5-(2x}{15}>= \frac{x+4)/3 ?

The solution to x/5-(2x}{15}>= \frac{x+4)/3 is x<=-5

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x/5-(2x}{15}>= \frac{x+4)/3

\frac{x}{5} - \frac{2 x}{15} \geq \frac{x + 4}{3}

x \leq - 5

Interval Notation

(- \infty, - 5]

Solution steps

\frac{x}{5} - \frac{2 x}{15} \geq \frac{x + 4}{3}

Multiply by LCM

x \geq 5 (x + 4)

Expand

5 (x + 4) : 5 x + 20

x \geq 5 x + 20

Move

5 x

to the left side

- 4 x \geq 20

Multiply both sides by

- 1

4 x \leq - 20

Divide both sides by

4

x \leq - 5

\frac{2}{5} x - 4 = - \frac{1}{5} x + 8

10 y^{2} - 21 y - 10

9 x - 10 = 5 x + 2 (2 x - 5)

\frac{6}{7} x = - 18

- 2 (7 - n) + 4

What is x/5-(2x}{15}>= \frac{x+4)/3 ?
The solution to x/5-(2x}{15}>= \frac{x+4)/3 is x<=-5