What is ((2x+1))/3 >= ((3x-2))/5 ?

The solution to ((2x+1))/3 >= ((3x-2))/5 is x>=-11

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((2x+1))/3 >= ((3x-2))/5

\frac{( 2 x + 1 )}{3} \geq \frac{( 3 x - 2 )}{5}

x \geq - 11

Interval Notation

[- 11, \infty)

Solution steps

\frac{( 2 x + 1 )}{3} \geq \frac{( 3 x - 2 )}{5}

Apply fraction cross multiply: if

\frac{a}{b} = \frac{c}{d}

then

a \cdot d = b \cdot c

(2 x + 1) \cdot 5 \geq 3 (3 x - 2)

Expand

(2 x + 1) \cdot 5 : 10 x + 5

Expand

3 (3 x - 2) : 9 x - 6

10 x + 5 \geq 9 x - 6

Move

5

to the right side

10 x \geq 9 x - 11

Move

9 x

to the left side

x \geq - 11

5 x - 1 < 3 x + 9

x, 15 \cdot x + 7.5 \cdot y + 5 \cdot z \leq 315

3 (x + 2) + 7 (x - 4) \geq 3 x - 1

a^{2} (x - 5) > b^{2} (x - 5)

\frac{( a + 3 )}{5} - \frac{( 2 a - 1 )}{2} > \frac{( 4 - a )}{2} - \frac{a}{10}

What is ((2x+1))/3 >= ((3x-2))/5 ?
The solution to ((2x+1))/3 >= ((3x-2))/5 is x>=-11