Question

Y+1 y+8 y+2 y+7

----- + ----- = ----- + -----
y+2 y+9 y+3 y+8

find y
plssssss help this is fraction

Answer 1

`(y+1)/(y+2)+(y+8)/(y+9)=(y+2)/(y+3)+(y+7)/(y+8)`

Write all fractions in terms of a common denominator:

`(y+1)/(y+2)=((y+1)(y+9)(y+3)(y+8))/((y+2)(y+9)(y+3)(y+8))`

`(y+8)/(y+9)=((y+8)^2(y+2)(y+3))/((y+2)(y+9)(y+3)(y+8))`

`(y+2)/(y+3)=((y+2)^2(y+9)(y+8))/((y+2)(y+9)(y+3)(y+8))`

`(y+7)/(y+8)=((y+7)(y+2)(y+9)(y+3))/((y+2)(y+9)(y+3)(y+8))`

Then move all fractions to one side and simplify the numerator:

The numerator dictates when the fraction reduces to 0. The denominator can never be 0, so we know that y cannot take any of the values -2, -9, -3, nor -8.

So the equation reduces to

`(y+1)(y+9)(y+3)(y+8)+(y+8)^2(y+2)(y+3)-(y+2)^2(y+9)(y+8)-(y+7)(y+2)(y+9)(y+3)=0`

Expand the left side; you would end up with

`-6(2y+11)=0`

`2y+11=0`

`2y=-11`

`\implies\boxed{y=-\frac{11}2}`