What is the answer to -(3y+12}{12}-\frac{2y-4)/4-(y+2)/8 =0 ?

The answer to -(3y+12}{12}-\frac{2y-4)/4-(y+2)/8 =0 is y=-2/7

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-(3y+12}{12}-\frac{2y-4)/4-(y+2)/8 =0

- \frac{3 y + 12}{12} - \frac{2 y - 4}{4} - \frac{y + 2}{8} = 0

y = - \frac{2}{7}

Decimal Notation

y = - 0.28571 \dots

Solution steps

- \frac{3 y + 12}{12} - \frac{2 y - 4}{4} - \frac{y + 2}{8} = 0

Multiply by LCM

- 6 (y + 4) - 12 (y - 2) - 3 (y + 2) = 0

Expand

- 6 (y + 4) - 12 (y - 2) - 3 (y + 2) : - 21 y - 6

- 21 y - 6 = 0

Move

6

to the right side

- 21 y = 6

Divide both sides by

- 21

y = - \frac{2}{7}

\frac{61}{28} + a = \frac{19}{4} a - \frac{11}{7}

(1 + 0.03)^{5} - 1

\frac{1}{x - 2} + \frac{2}{3 x - 21} \geq 0

x^{2} - 10 x = - 21

2 x - \frac{1}{3} \leq 8

What is the answer to -(3y+12}{12}-\frac{2y-4)/4-(y+2)/8 =0 ?
The answer to -(3y+12}{12}-\frac{2y-4)/4-(y+2)/8 =0 is y=-2/7
Solve for y: -(3y+12}{12}-\frac{2y-4)/4-(y+2)/8 =0
The solution is y=-2/7