Question

question:- solve for x for which DE II AB​

Answer 1

Answer:

2

Step-by-step explanation:

$\frac{x + 3}{x + 3 + 3x + 19} = \frac{x}{x + 3x + 4}$ ----------------------- Basic Proportionality Theorem (∵ DE II AB​)

⇒ $\frac{x + 3}{4x + 22} = \frac{x}{4x + 4}$

⇒ (x + 3)(4x + 4) = (4x + 22)x ---------- Cross multiplication

⇒ $4x^{2}$ + 4x + 12x + 12 = $4x^{2}$ + 22x

⇒ $4x^{2}$ - $4x^{2}$ = 22x - 4x - 12x - 12

⇒ 22x - 16x - 12 = 0

⇒ 6x - 12 = 0

6x = 12

x = 2

Hope it helps