Question

I need help with 5 and 6 the entire thing is one question just sectional

Answer 1

Given:

The expression is,

`(1)/(2)+(15)/(2x-14)-(3x-15)/(x^2-7x)`

Step-by-step explanation:

Simplify the expression.

`\begin{gathered} (1)/(2)+(15)/(2x-14)-(3x-15)/(x^2-7x)=(1)/(2)+(15)/(2(x-7))-\frac{3x-15}{x^{}(x-7)} \\ =(x(x-7)+15\cdot x-(3x-15)\cdot2)/(2x(x-7)) \\ =(x^2-7x+15x-6x+30)/(2x(x-7)) \\ =(x^2+2x+30)/(2x(x-7)) \end{gathered}`

The function is not defined at which denominator is equal to 0. So,

`\begin{gathered} 2x(x-7)=0 \\ x=0,7 \end{gathered}`

Thus function is defined for all values of x except 0 and 7. So domain of the function is,

`(-\infty,0)\cup(0,7)\cup(7,\infty)`