How do you solve $6 y + 3 (y - 9) = 4 (y + 1) - 2$ $6y + 3( y - 9) = 4( y + 1) - 2$ ?

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Answer 1 · 2018-06-28T05:26:22

$y = \frac{29}{5}$ $y=29/5$

Given -

$6 y + 3 (y - 9) = 4 (y + 1) - 2$ $6y+3(y-9)=4(y+1)-2$

Remove the brackets

$6 y + 3 y - 27 = 4 y + 4 - 2$ $6y+3y-27=4y+4-2$

Simplify

$9 y - 27 = 4 y + 2$ $9y-27=4y+2$

Add $- 4 y$ $-4y$ on both sides

$9 y - 27 - 4 y = 4 y + 2 - 4 y$ $9y-27-4y=4y+2-4y$

$5y-27=cancel(4y)+2cancel(-4y)$

$5y-27=2$

Add $+27$ to both sides

$5y-27+27=2+27$

$5ycancel(-27)cancel(+27)=2+27$

$5y=29$

$y=29/5$

How do you solve 6y + 3( y - 9) = 4( y + 1) - 26y+3(y−9)=4(y+1)−26y + 3( y - 9) = 4( y + 1) - 2?