Answer:
To find the solution to the system of equations \( \frac{2}{x+2} + \frac{1}{10} = \frac{3}{x+3} \), we need to solve for \(x\).
Multiplying each term by \(10(x + 2)(x + 3)\) to clear the denominators:
\[10(x + 3) + (x + 2)(x + 3) = 3(10)(x + 2)\]
\[10(x + 3) + (x^2 + 5x + 6) = 30x + 60\]
\[10x + 30 + x^2 + 5x + 6 = 30x + 60\]
\[x^2 + 15x + 36 = 30x + 60\]
\[x^2 - 15x - 24 = 0\]
Now, we can solve for \(x\) using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[x = \frac{15 \pm \sqrt{(-15)^2 - 4(1)(-24)}}{2(1)}\]
\[x = \frac{15 \pm \sqrt{225 + 96}}{2}\]
\[x = \frac{15 \pm \sqrt{321}}{2}\]
Since the discriminant (\(b^2 - 4ac\)) is positive, there are two real solutions. Therefore, the correct statement is:
1) The system of equations has TWO valid solutions and NO extraneous solutions.
