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Find y if \(\dfrac{y^2 - 9y + 8}{y-1} + \dfrac{3y^2 +16y-12 }{3y -2} = -3\)

 Apr 21, 2019
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Solve for y:
(y^2 - 9 y + 8)/(y - 1) + (3 y^2 + 16 y - 12)/(3 y - 2) = -3

Simplify and substitute x = y - 1.
(y^2 - 9 y + 8)/(y - 1) + (3 y^2 + 16 y - 12)/(3 y - 2) = 2 (y - 1)
 = 2 x:
2 x = -3

Divide both sides by 2:
x = -3/2

Substitute back for x = y - 1:
y - 1 = -3/2

Add 1 to both sides:
y = -1/2

 Apr 21, 2019

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