The solution set for the given equation is x=-1 .
Solution set is written as {-1}
for better understanding check the calculation below.
Calculation :
We need to solve the given equation
[tex]\frac{x+3}{x+2} =3+\frac{1}{x}[/tex]
To solve any equation we take LCD.
LCD of the given equation is x(x+2)
Multiply LCD with each term to remove the denominator
[tex]\frac{x+3}{x+2}\cdot x(x+2) =3\cdot x(x+2)+\frac{1}{x}\cdot x(x+2)\\x(x+3)=3x(x+2)+x+2\\x^2+3x=3x^2+6x+x+2\\[/tex]
Subtract x^2 and 3x from both sides of the equation to get 0 on the left side of the equation
[tex]0=3x^2+6x+x+2-x^2-3x\\0=2x^2+4x+2\\[/tex]
Now divide the whole equation by 2 and then we factor it
[tex]0=x^2+2x+1\\0=(x+1)(x+1) \\x+1=0\\x=-1[/tex]
The solution set for the given equation is x=-1
Solution set is written as {-1}
Learn more information about the ' rational equaion ' here : brainly.com/question/8519709
