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#