Let $f(x)=6 x+5$ and $g(x)=4 x-7$. Find $(f+g)(x),(f-g)(x),(f g)(x)$, and $\left(\frac{f}{g}\right)(x)$. Give the domain of $\frac{f}{g}$ $(f+g)(x)=\square$ (Simplify your answer.) $(f-g)(x)=\square$ (Simplify your answer) $(f g)(x)=\square($ Simplify your answer.) $\left(\frac{f}{g}\right)(x)=\square($ Simplify your answer.) The domain of $\frac{f}{g}$ is

Step 1 ：Define the functions $f(x)=6x+5$ and $g(x)=4x-7$.

Step 2 ：To find $(f+g)(x)$, add the functions $f(x)$ and $g(x)$ together to get $10x - 2$.

Step 3 ：To find $(f-g)(x)$, subtract $g(x)$ from $f(x)$ to get $2x + 12$.

Step 4 ：To find $(fg)(x)$, multiply $f(x)$ and $g(x)$ together to get $24x^2 - 2x - 35$.

Step 5 ：To find $\left(\frac{f}{g}\right)(x)$, divide $f(x)$ by $g(x)$ to get $\frac{6x + 5}{4x - 7}$.

Step 6 ：The domain of $\frac{f}{g}$ is all real numbers except for the values of $x$ that make $g(x) = 0$, because division by zero is undefined. Solve $4x - 7 = 0$ to find that $x \neq \frac{7}{4}$.

Step 7 ：Final Answer: $(f+g)(x)=\boxed{10x - 2}$, $(f-g)(x)=\boxed{2x + 12}$, $(f g)(x)=\boxed{24x^2 - 2x - 35}$, $\left(\frac{f}{g}\right)(x)=\boxed{\frac{6x + 5}{4x - 7}}$, The domain of $\frac{f}{g}$ is $\boxed{x \neq \frac{7}{4}}$