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

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Answer 1 · 2018-04-15T16:25:11

$y=cancel1, y=4$

common denominator
$(3/3)(4/(y-1)) + ((y-1)/(y-1))(2/3) = (3/3)(6/(y-1))$

$12/(3(y-1)) + (2(y-1))/(3(y-1)) = 18/(3(y-1))$

$(12+2y-2)/(3(y-1)) = 18/(3(y-1))$

$(2y + 10)/(3(y-1)) = 18/(3(y-1))$

$(2y+10)(3y-3) = 18(3y-3)$

$6y^2 + 24y -30 = 54y - 54$

$6y^2 +24y- 54y -30 +54 = 0$

$6y^2 - 30y +24 = 0$

$(3y-3)(2y-8)$

$y=1, y=4$

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