What is (d-3}{10}>= \frac{2d+3)/5+(d+3)/3 ?

The solution to (d-3}{10}>= \frac{2d+3)/5+(d+3)/3 is d<=-3

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(d-3}{10}>= \frac{2d+3)/5+(d+3)/3

\frac{d - 3}{10} \geq \frac{2 d + 3}{5} + \frac{d + 3}{3}

d \leq - 3

Interval Notation

(- \infty, - 3]

Solution steps

\frac{d - 3}{10} \geq \frac{2 d + 3}{5} + \frac{d + 3}{3}

Multiply by LCM

3 (d - 3) \geq 6 (2 d + 3) + 10 (d + 3)

Expand

3 (d - 3) : 3 d - 9

Expand

6 (2 d + 3) + 10 (d + 3) : 22 d + 48

3 d - 9 \geq 22 d + 48

Move

9

to the right side

3 d \geq 22 d + 57

Move

22 d

to the left side

- 19 d \geq 57

Multiply both sides by

- 1

19 d \leq - 57

Divide both sides by

19

d \leq - 3

z^{5} \cdot z^{6}

∣ - 2 - 5 x ∣ = \frac{7}{4}

64 - y^{4}

2 x^{3} - 5 x^{2} - 2 x + 2

(\frac{a ^{- 3}}{b ^{4}})^{\frac{1}{12}}

What is (d-3}{10}>= \frac{2d+3)/5+(d+3)/3 ?
The solution to (d-3}{10}>= \frac{2d+3)/5+(d+3)/3 is d<=-3