Answer:
Here we need to solve:
[tex]\frac{2}{y - 3} + \frac{6}{y + 3 } = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} }[/tex]
The sum of the fractions is equal to the quotient between the fractions.
Notice that the two values:
y = 3
y = -3
make the denominator equal to zero, so those values are restricted.
We can simplify the right side to get:
[tex]\frac{2}{y - 3} + \frac{6}{y + 3 } = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} } = \frac{2*(y + 3)}{6*(y - 3)} = 3*\frac{y + 3}{y - 3}[/tex]
Now we can multiply both sides by (y - 3)
[tex](y - 3)*(\frac{2}{y - 3} + \frac{6}{y + 3 }) = 3*(y + 3)\\2 + 6*\frac{y -3}{y + 3} = 3*(y + 3)[/tex]
Now we can multiply both sides by (y + 3)
[tex](2 + 6*\frac{y -3}{y + 3})*(y + 3) = 3*(y + 3)*(y + 3)[/tex]
[tex]2*(y + 3) + 6*(y - 3) = 3*(y + 3)*(y + 3)\\\\2*y + 6 + 6*y - 18 = 3*(y^2 + 2*y*3 + 9)\\\\8*y - 12 = 3*y^2 + 6*y + 33\\\\0 = 3*y^2 + 6*y + 33 - 8*y + 12\\\\0 = 3*y^2 - 2*y + 45[/tex]
First, let's see the determinant of that quadratic equation:
[tex]D = (-2)^2 - 4*3*45 = -536[/tex]
We can see that it is negative, thus, there are no real solutions of the equation.
Thus, there is no value of y such that the origina equation is true,
