Question

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

Answer 1

Answer:

{-2, 1}

Step-by-step explanation:

$\frac{y+4}{2y} +\frac{y-2}{3} =\frac{3y^2+10}{6y}$

Make the fractions have a common denominator.

$\frac{3}{3} (\frac{y+4}{2y}) +\frac{x}{y} (\frac{y-2}{3}) =\frac{3y^2+10}{6y}\\\frac{3y+12}{6y} +\frac{2y^2-4y}{6y} =\frac{3y^2+10}{6y}\\\frac{2y^2-y+12}{6y} =\frac{3y^2+10}{6y}$

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}