We have to solve the equation
[tex]\frac{5}{x+8}=\frac{3}{x}[/tex]
In order to solve it for x, subtract 3/x to both sides
[tex]\frac{5}{x+8}-\frac{3}{x}=0[/tex]
[tex]\mathrm{Find\:Least\:Common\:Multiplier\:of\:}x+8,\:x:\quad x\left(x+8\right)[/tex]
Multiply both sides by LCM
[tex]\frac{5}{x+8}x\left(x+8\right)-\frac{3}{x}x\left(x+8\right)=0\cdot \:x\left(x+8\right)[/tex]
On simplifying, we get
[tex]5x-3x-24=0\\2x-24=0\\2x=24\\x=12[/tex]
To check the extraneous solution, we plugging this value of x in the given equation and check if it satisfies or not
[tex]\frac{5}{12+8}=\frac{3}{12}[/tex]
[tex]\frac{5}{20}=\frac{1}{4}[/tex]
[tex]\frac{1}{4}=\frac{1}{4}[/tex]
This is true. Hence, x=12 is a solution (non-extraneous) to the given equation.
A is the correct option.