Question

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Answer 1

Answer:

$x = 3$

Step-by-step explanation:

$In ΔABC, \\ DE \: II \: BC \\\frac{AD}{DB} = \frac{AE}{EC} (BPT \: Theorem) \\ \frac{x}{x + 1} = \frac{x + 3}{x + 5} \\ x(x + 5) = (x + 1)(x + 3) \\ x(x + 5) = x(x + 3) + 1(x + 3) \\ {x}^{2} + 5x = {x}^{2} + 3x + x + 3 \\ {x}^{2} - {x}^{2} + 5x - 3x - x - 3 = 0 \\ x - 3 = 0 \\ x = 3$