Solving the equation we have:
[tex]\begin{gathered} \frac{x+3}{x-3}=\frac{12}{3} \\ \frac{x+3}{x-3}=4\text{ (Simplifying the fraction)} \\ x+3=4(x-3)\text{ (Multiplying x-3 on both sides of the equation)} \\ x+3=4x-12\text{ (Distributing)} \\ x+3+12=4x\text{ (Adding 12 to both sides of the equation)} \\ 3+12=4x-x\text{ (Subtracting x from both sides of the equation)} \\ 15=3x\text{ (Adding)} \\ \frac{15}{3}=x\text{ (Dividing by 3 on both sides of the equation)} \\ 5=x\text{ } \end{gathered}[/tex]
The solution is x=5 and it is valid as the result of replacing it in the denominator is not zero. ( 5 - 3 ≠ 0)
