solve eqn. 4/(y-1) + 2/(y-3) = 6/(y-2).?

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

the first thing to notice is that y cannot be 1, 2, or 3, otherwise you will have a denominator equal to zero.

The LCD will be the product of the three denominators. Multiply both sides by this LCD to get rid of the fractions. We now have:

4(y - 3)(y - 2) + 2(y - 1)(y - 2) = 6(y - 1)(y - 3)

Now simplify into a quadratic, then solve:

4(y² - 5y + 6) + 2(y² - 3y + 2) = 6(y² - 4y + 3)

4y² - 20y + 24 + 2y² - 6y + 4 = 6y² - 24y + 18

6y² - 26y + 28 = 6y² - 24y + 18

Looks like the y² terms will now cancel out, so we just have a linear equation:

-26y + 28 = -24y + 18

-2y = -10

y = 5

That is not in the list of invalid solutions, so that is your answer. As a sanity test, let's do the math:

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

4 / (5 - 1) + 2 / (5 - 3) = 6 / (5 - 2)

4 / 4 + 2 / 2 = 6 / 3

1 + 1 = 2

2 = 2

TRUE

So it is correct. Again, the answer is:

y = 5

Perhaps your problem is the mistake in

(4)(y-3)(y-2) + 2(y-1)(y-2) = 6(y-1)(y-2) which should be

(4)(y-3)(y-2) + 2(y-1)(y-2) = 6(y-1)(y-3) --The last number is 3 not 2.

4 (y - 3) (y - 2) + 2 (y - 1) (y - 2) = 6 (y - 1) (y - 3)

4 ( y² - 5y + 6 ) + 2 (y² - 3y + 2 ) = 6 (y² - 4y + 3 )

6y² - 26y + 28 = 6y² - 24y + 18

10 = 2y

y = 5

you have to convert to the same number in both denominators first -

to convert you subtract -1 from both sides --1=+1 so -3--1=-3+1=2