Question

In AABC, DE parallel to BC
AD=8x+9
DB=x+3
AE = 3x+4
Ec = x
Find the value of 'x'
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Answer 1

Given,DE//AB

$\frac{AD}{DC} = \frac{BE}{EC} \\ \\ \frac{8x + 9}{x + 3} = \frac{3x + 4}{x} \\ \\ {8x}^{2} +9x= {3x}^{2} +13x+12 \\ \\ {5x}^{2} −4x−12=0 \\ \\ {5x}^{2} −10x+6x−12=0 \\ \\ 5x(x−2)+6(x−2)=0 \\ \\ (5x+6)(x−2)=0 \\ \\ x = \frac{ - 6}{5} \: and \: 2$

Therefore x is 2.