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

Answer 1

Genesis Allen

#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#

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Answer 2

Julia Adams

To solve the equation 4/(y-1) + 2/3 = 6/(y-1), we can start by getting rid of the denominators. Multiply both sides of the equation by (y-1) to eliminate the denominators. This gives us 4 + (2/3)(y-1) = 6. Next, distribute the (2/3) to both terms inside the parentheses: 4 + (2/3)y - 2/3 = 6. Combine like terms by adding 4 and -2/3: (2/3)y + 10/3 = 6. Subtract 10/3 from both sides: (2/3)y = 6 - 10/3. Simplify the right side: (2/3)y = 18/3 - 10/3 = 8/3. Finally, multiply both sides by 3/2 to isolate y: y = (8/3)(3/2) = 4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.