Question

in figure DE is parallel to BC if BD =x minus 3 ab equals to 2 x CE= 2 x minus 2 and ac is equals to 2 X + 3 find x

Answer 1

Answer:

WHERE IS THE FIGURE BRO?????...

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Answer 2

The required equation for x is $x^2-4x+9=0$

Step-by-step explanation:

Since we have given that

DE is parallel to BC,

And

$BD=x-3\\\\AB=2x\\\\CE=2x-2\\\\AC=2x+3$

Using "Basic proportionality theorem", we get that

$\dfrac{BD}{AB}=\dfrac{CE}{AC}\\\\\dfrac{x-3}{2x}=\dfrac{2(x-2)}{2(x+3)}\\\\(x-3)(2x+6)=(2x-4)(2x)\\\\2x^2+6x-6x-18=4x^2-8x\\\\-18=4x^2-2x^2-8x\\\\-18=2x^2-8x\\\\2x^2-8x+18=0\\\\x^2-4x+9=0$

Hence, the required equation for x is $x^2-4x+9=0$

# learn more:

In the figure of triangle ABC, DE parallel AB. if AD=2 x, DC= x + 3, BE is equal to 2 x- 1 and CE=x ,then find the value of x

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