How do you solve $- 3 y + 4 = - 1$ $-3y+4=-1$ ?

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Answer 1 · 2017-01-06T04:27:30

$y = 1 \frac{2}{3}$ $y=1 2/3$

$- 3 y + 4 = - 1$ $-3y+4=-1$

Subtract $4$ $4$ from each side.

$- 3 y = - 5$ $-3y=-5$

Divide both sides by $- 3$ $-3$ .

$y = \frac{5}{3} = 1 \frac{2}{3}$ $y=5/3=1 2/3$

How do you solve -3y+4=-1−3y+4=−1-3y+4=-1?