What is (3x-1)/7-(2-4x)/3 >= (-5-14)/(14)+7x+6 ?

The solution to (3x-1)/7-(2-4x)/3 >= (-5-14)/(14)+7x+6 is x<=-229/220

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(3x-1)/7-(2-4x)/3 >= (-5-14)/(14)+7x+6

\frac{3 x - 1}{7} - \frac{2 - 4 x}{3} \geq \frac{- 5 - 14}{14} + 7 x + 6

x \leq - \frac{229}{220}

Interval Notation

(- \infty, - \frac{229}{220}]

Decimal Notation

x \leq - 1.04090 \dots

Solution steps

\frac{3 x - 1}{7} - \frac{2 - 4 x}{3} \geq \frac{- 5 - 14}{14} + 7 x + 6

\frac{- 5 - 14}{14} = - \frac{19}{14}

\frac{3 x - 1}{7} - \frac{2 - 4 x}{3} \geq - \frac{19}{14} + 7 x + 6

Combine the fractions

6 - \frac{19}{14} : \frac{65}{14}

\frac{3 x - 1}{7} - \frac{2 - 4 x}{3} \geq 7 x + \frac{65}{14}

Multiply by LCM

6 (3 x - 1) - 14 (- 4 x + 2) \geq 294 x + 195

Expand

6 (3 x - 1) - 14 (- 4 x + 2) : 74 x - 34

74 x - 34 \geq 294 x + 195

Move

34

to the right side

74 x \geq 294 x + 229

Move

294 x

to the left side

- 220 x \geq 229

Multiply both sides by

- 1

220 x \leq - 229

Divide both sides by

220

x \leq - \frac{229}{220}

0.5 (2.8 x + 1.7) = 0.43

3 (\frac{1}{3})^{2} - \frac{1}{3}

1 \cdot 1 0^{- 3} = \frac{\frac{x}{506.4}}{0.025}

\frac{2 π}{3} + 2 π

\frac{9 !}{( 9 - 7 ) ! 7 !}

What is (3x-1)/7-(2-4x)/3 >= (-5-14)/(14)+7x+6 ?
The solution to (3x-1)/7-(2-4x)/3 >= (-5-14)/(14)+7x+6 is x<=-229/220