Question

If DE || BC, AD = 2x, DC = x+3, BE = 2x-1 & CE = x then find the value of x?

Help plzzzzzzz... ​

Answer 1

Answer:

there should be a diagram plz send with diagrams

Answer 2

AD = 2x ,DC = x + 3, BE =2x - 1 ,CE = x

Using Thales Theorem

CD/AD = CE/BE

(i.e.,)

$(x + 3) \div2x = (x) \div (2x - 1) \\ 5x = 3 \\ x = 5 \div 3$

So, x=5/3

calculation part:

(x+3)/2x = x/(2x-1)

(x+3)(2x-1) = x (2x)

2x^2 -x +6x -3 = 2x^2

2x^2 +5x -3 -2x^2 = 0

5x+3 = 0

x = 5/3