Question
What values of $x$ will make $DE || AB$ in the given figure$?$
✓
Answer
In the figure, in $\triangle\text{ABC},$
$DE || AB$
$\therefore\frac{\text{AD}}{\text{DC}}=\frac{\text{BE}}{\text{EC}}$
$\Rightarrow\frac{3\text{x}+19}{\text{x}+3}=\frac{3\text{x}+4}{\text{x} }$
$⇒ x(3x + 19) = (x + 3)(3x + 4)$
$ \Rightarrow 3 x^2+19 x=3 x^2+4 x+9 x+12 $
$ \Rightarrow 3 x^2+19 x-3 x^2-4 x-9 x=12$
$\Rightarrow6\text{x}=12\Rightarrow\text{x}=\frac{12}{6}=2$
$\therefore x = 2$
$DE || AB$
$\therefore\frac{\text{AD}}{\text{DC}}=\frac{\text{BE}}{\text{EC}}$
$\Rightarrow\frac{3\text{x}+19}{\text{x}+3}=\frac{3\text{x}+4}{\text{x} }$
$⇒ x(3x + 19) = (x + 3)(3x + 4)$
$ \Rightarrow 3 x^2+19 x=3 x^2+4 x+9 x+12 $
$ \Rightarrow 3 x^2+19 x-3 x^2-4 x-9 x=12$
$\Rightarrow6\text{x}=12\Rightarrow\text{x}=\frac{12}{6}=2$
$\therefore x = 2$
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