Answer:
{-2, 1}
Step-by-step explanation:
[tex]\frac{y+4}{2y} +\frac{y-2}{3} =\frac{3y^2+10}{6y}[/tex]
Make the fractions have a common denominator.
[tex]\frac{3}{3} (\frac{y+4}{2y}) +\frac{x}{y} (\frac{y-2}{3}) =\frac{3y^2+10}{6y}\\\frac{3y+12}{6y} +\frac{2y^2-4y}{6y} =\frac{3y^2+10}{6y}\\\frac{2y^2-y+12}{6y} =\frac{3y^2+10}{6y}[/tex]
Now that both sides of the equation have a common denominator, you can cancel out the denominators.
[tex]2y^2-y+12=3y^2+10[/tex]
Now, set the equation equal to zero and factor.
[tex]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[/tex]
The y-values are {-2, 1}
