In a triangle abc de parallel to bc find the value of x

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As DE\xa0{tex}\\parallel{/tex} BC{tex}\\therefore \\frac{AD}{AB}=\\frac{AE}{AC}{/tex},\xa0{tex}{/tex}.{tex}\\frac{x}{2x+1}=\\frac{x+3}{2x+8}{/tex}(x + 3)(2x + 1) = x(2x + 8)2×2 + x + 6x + 3 = 2×2 + 8×3 = 8x – 7xx = 3