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

1.	In a triangle abc de parallel to bc find the value of x
Answer» 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)2x2 + x + 6x + 3 = 2x2 + 8x3 = 8x - 7xx = 3

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