3y-5/y 2. 4b/2x(6x-5) 3. 7/x(7-2x) 4. x2-32/(x+5)(x-6) 5. If y=5x+4/2x+3 a. For what value of x is y not defined? b. Determine the value of x for which y Vanishes c. For what range of values of x is y>3?

Set the denominator to zero and solve for $x$.

The value of $x$ for which $y$ is not defined is $\boxed{-\frac{3}{2}}$.

Set the numerator to zero and solve for $x$.

The value of $x$ for which $y$ vanishes is $\boxed{-\frac{4}{5}}$.

Numerator: $-x-5 = 0 \implies x = -5$
Denominator: $2x+3 = 0 \implies x = -\frac{3}{2}$

The critical points divide the number line into three intervals: $(-\infty, -5)$, $(-5, -\frac{3}{2})$, and $(-\frac{3}{2}, \infty)$.

• For $x < -5$ (e.g., $x=-6$): Numerator: $-(-6)-5 = 6-5 = 1$ (positive) Denominator: $2(-6)+3 = -12+3 = -9$ (negative) Fraction: $\frac{\text{positive}}{\text{negative}} = \text{negative}$. So, $\frac{-x-5}{2x+3} < 0$.

• For $-5 < x < -\frac{3}{2}$ (e.g., $x=-2$): Numerator: $-(-2)-5 = 2-5 = -3$ (negative) Denominator: $2(-2)+3 = -4+3 = -1$ (negative) Fraction: $\frac{\text{negative}}{\text{negative}} = \text{positive}$. So, $\frac{-x-5}{2x+3} > 0$.

• For $x > -\frac{3}{2}$ (e.g., $x=0$): Numerator: $-0-5 = -5$ (negative) Denominator: $2(0)+3 = 3$ (positive) Fraction: $\frac{\text{negative}}{\text{positive}} = \text{negative}$. So, $\frac{-x-5}{2x+3} < 0$.

The range of values for $x$ for which $y > 3$ is $\boxed*{-5 < x < -\frac{3}{2}*}$.