Question

Please help me with this quickly, make it simple and easy, thank you

Answer 1

`(x+5)/(x^2-x)-(3)/(x)=(1)/(x-1)`

Since neither of the denominators of the equation can be equal to 0, we know the domain is going to be all numbers different from 0 and 1.

`\begin{gathered} denominators: \\ x=0 \\ x^2-x=0 \\ x-1=0 \\ x=1 \end{gathered}`

`(x+5)/(x^2-x)-(3)/(x)=(1)/(x-1);x\\e1;x\\e0`

Then, solve the sum of fractions on the left leaving the denominator as a factors

`\begin{gathered} (x(x+5)-3(x^2-x))/(x(x^2-x))=(1)/(x-1) \\ \end{gathered}`

factor the denominator and simplify

`(x^2+5x-3x^2+3x)/(x* x(x-1))=(1)/(x-1)`

multiply both sides by x-1

`\begin{gathered} (-2x^2+8x)/(x^2)=1\frac{}{} \\ \end{gathered}`

separate the fraction on the left side

`-2+(8)/(x)=1`

solve for x,

`\begin{gathered} (8)/(x)=1+2 \\ (8)/(x)=3 \\ 8=3x \\ x=(3)/(8) \end{gathered}`

Answer:

x is equal to 3/8