What is 6(3x-1)-5(4x-3)>= 9-2x ?

The solution to 6(3x-1)-5(4x-3)>= 9-2x is True for all x

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6(3x-1)-5(4x-3)>= 9-2x

6 (3 x - 1) - 5 (4 x - 3) \geq 9 - 2 x

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

6 (3 x - 1) - 5 (4 x - 3) \geq 9 - 2 x

Expand

6 (3 x - 1) - 5 (4 x - 3) : - 2 x + 9

- 2 x + 9 \geq 9 - 2 x

Move

9

to the right side

- 2 x \geq - 2 x

Move

2 x

to the left side

0 \geq 0

Therefore, the solution is

True for all x

x < \frac{- 3}{2}

4 - 3 (x - 2) < 3 + 2 (x + 6)

8 + 4 x > 6

\frac{( 5 x - 2 )}{3} - \frac{( x - 8 )}{4} \geq \frac{( x + 14 )}{2}

\frac{- 1}{5} < \frac{x + ( - 1 )}{4}

What is 6(3x-1)-5(4x-3)>= 9-2x ?
The solution to 6(3x-1)-5(4x-3)>= 9-2x is True for all x