How do you solve $3 y (4 y - 1) - 2 y (6 y - 5) = 9 y - 8 (3 + y)$ $3y ( 4y - 1) - 2y ( 6y - 5) = 9y - 8( 3+ y )$ ?

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Answer 1 · 2017-01-03T22:31:50

$y = - 4$ $y = -4$

First distribute everything:

$12 y^{2} - 3 y - 12 y^{2} + 10 y = 9 y - 24 - 8 y$ $12y^2-3y-12y^2 + 10y = 9y-24-8y$

Then combine like terms:

$7 y = y - 24 \to$ $7y = y-24 ->$ the two $12 y^{2}$ $12y^2$ terms cancel out

You can then subtract $y$ $y$ from both sides:

$6 y = - 24$ $6y = -24$

Then divide both sides by $- 14$ $-14$ to get that

$y = \frac{- 24}{6} = - 4$ $y = (-24)/6 = -4$

How do you solve 3y(4y−1)−2y(6y−5)=9y−8(3+y)3y ( 4y - 1) - 2y ( 6y - 5) = 9y - 8( 3+ y )?