What is the answer to (3x+1)/7-(2-4x)/3 =(5x-4}{14}+\frac{7x)/6 ?

The answer to (3x+1)/7-(2-4x)/3 =(5x-4}{14}+\frac{7x)/6 is x=1

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(3x+1)/7-(2-4x)/3 =(5x-4}{14}+\frac{7x)/6

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

x = 1

Solution steps

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

Multiply by LCM

6 (3 x + 1) - 14 (- 4 x + 2) = 3 (5 x - 4) + 49 x

Expand

6 (3 x + 1) - 14 (- 4 x + 2) : 74 x - 22

Expand

3 (5 x - 4) + 49 x : 64 x - 12

74 x - 22 = 64 x - 12

Move

22

to the right side

74 x = 64 x + 10

Move

64 x

to the left side

10 x = 10

Divide both sides by

10

x = 1

(4 + i) - (3 - 5 i) (- 2 + 5 i)

5 x < 10

x (x - 2)^{2} > 0

lo g_{7} (0)

\frac{e ^{0}}{2}

What is the answer to (3x+1)/7-(2-4x)/3 =(5x-4}{14}+\frac{7x)/6 ?
The answer to (3x+1)/7-(2-4x)/3 =(5x-4}{14}+\frac{7x)/6 is x=1
Solve for x: (3x+1)/7-(2-4x)/3 =(5x-4}{14}+\frac{7x)/6
The solution is x=1