Question

How do you solve `\frac { 2} { y - 2} + \frac { 4} { y - 3} = \frac { 1} { y ^ { 2} - 5y + 6}`?

Answer 1

`y=5/2`

Explanation:

Given:

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

Note that:

`y^2-5y+6 = (y-2)(y-3)`

So, multiplying both sides of the equation by `y^2-5y+6` we get:

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

Multiplying out, this becomes:

`2y-6+4y-8=1`

Hence:

`6y=15`

So:

`y = 15/6 = 5/2`

Answer 2

Given: `2/(y - 2) + 4/(y - 3) = 1/(y^2 - 5y + 6)`

Please observe that the quadratic factors into the same factors as the other denominators:

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

If we multiply both sides by `(y-2)(y-3)`, we will remove all of the denominators:

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

Use the distributive property:

`2y-6+4y-8=1`

Combine like terms:

`6y=15`

`y = 5/2 larr` answer

check:

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

`4 - 8 = 1/(-1/4)`

`-4 = -4`

This checks.