Question

In ∆ABC,DE || AB, AD=8x+9,CD=x+3BE=3x+4,CE=x Find the value of x​

Answer 1

Answer:

hlo mate here is the answer

Answer 2

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.