Question

How do you solve `\frac { 9y } { 3y + 9} + \frac { 5y - 13} { 2y + 6} = \frac { 8y + 1} { y + 3}`?

Answer 1

There is no solution.

Explanation:

`(9y)/(3y+9) +(5y-13)/(2y+6) = (8y+1)/(y+3)`

Observe that we cannot have `3y+9=0`, `2y+6=0` and `y+3=0`, as they appear in denominator i.e. `y!=-3`.

Now factorising the denominators:

`(cancel9^3y)/(cancel3(y+3)) +(5y-13)/(2(y+3)) = (8y+1)/(y+3)`

Get rid of the denominators by multiplying each term by the LCD (which in this case is `color(blue)(2(y+3))`) and cancelling the like factors

`(color(blue)(2cancel((y+3)))xx3y)/cancel((y+3))+(color(blue)cancel((2(y+3)))xx(5y-13))/cancel(2(y+3)) = (color(blue)(2cancel((y+3)))xx(8y+1))/cancel((y+3))`

This leaves us with:

`6y +5y-13 = 16y+2`

`-13-2 = 16y -6y-5y`

`-15 = 5y`

`y = -3`

But as we cannot have `y = -3`, there is no solution.