Question

(y + 4 / 2y) + (y-2 / 3) = (3y^2 + 10 / 6y) for y
{-2, 1}
{-2, 3}
{}

Answer 1

{-2, 1}

Explanation:

`(y+4)/(2y) +(y-2)/(3) =(3y^2+10)/(6y)`

Make the fractions have a common denominator.

`(3)/(3) ((y+4)/(2y)) +(x)/(y) ((y-2)/(3)) =(3y^2+10)/(6y)\\(3y+12)/(6y) +(2y^2-4y)/(6y) =(3y^2+10)/(6y)\\(2y^2-y+12)/(6y) =(3y^2+10)/(6y)`

Now that both sides of the equation have a common denominator, you can cancel out the denominators.

`2y^2-y+12=3y^2+10`

Now, set the equation equal to zero and factor.

`2y^2-y+12=3y^2+10\\-y^2-y+2=0\\(-y-2)(y-1)=0\\\\-y-2=0\\y=2\\\\y-1=0\\y=-1`

The y-values are {-2, 1}