Question

Someone please help? 25 points!!!!​

Answer 1

Use basic proportionality theorem

$\\ \sf\longmapsto \dfrac{AD}{DC}=\dfrac{BE}{AC}$

$\\ \sf\longmapsto \dfrac{2x}{x+3}=\dfrac{2x-1}{x}$

• Use cross multiplication

$\\ \sf\longmapsto 2x^2=(2x-1)(x+3)$

$\\ \sf\longmapsto 2x^2=2x(x+3)-1(x+3)$

$\\ \sf\longmapsto 2x^2=2x^2+6x-x-3$

$\\ \sf\longmapsto 2x^2-2x^2=5x-3$

$\\ \sf\longmapsto 5x-3=0$

$\\ \sf\longmapsto 5x=3$

$\\ \sf\longmapsto x=\dfrac{3}{5}$

Answer 2

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

• Find the value of x.

$\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}$

Given,

• DE || AB
• AD = 2x
• CD = x + 3
• BE = 2x - 1
• CE = x

Value of x (CE) = ?

__________________

We can solve this question using the Basic Proportionality Theorem (BPT). According to BPT, if a line is drawn parallel (||) to 1 side of a triangle to intersect the other 2 sides in 2 distinct points then the other 2 sides will be divided in the same ratio.

__________________

So in this triangle,

$\tt\frac{CD}{AD} = \frac{CE}{BE}$

Then,

$\tt \frac{x + 3}{2x} = \frac{x}{2x - 1} \\ \\ \sf \: Using \: cross \: multiplication.... \\ \\ \tt \: x + 3(2x - 1) = x(2x) \\ \tt \: x(2x - 1) + 3(2x - 1) = {2x}^{2} \\ \tt \: {2x}^{2} - x + 6x - 3 = {2x}^{2} \\ \\ \sf Cancel \: out \: 2 {x}^{2} \: from \: both \: the \: sides \\ \\ \tt \: - x + 6x - 3 = 0 \\ \tt \: - x + 6x = 3 \\ \tt \: 5x = 3 \\ \large\boxed{\boxed{\bf \: x = \frac{3}{5} }}$

__________________

• The value of x will be 3/5 (option b.)