Respuesta :

Triangles JFE and JLK are similar provided that line segments FE and LK are parallel. This means that corresponding parts of these triangles occur proportionally to one another. In particular,

[tex]\dfrac{JF}{JL}=\dfrac{JE}{JK}=\dfrac{FE}{LK}[/tex]

From this relation we get

[tex]\dfrac{JF}{JL}=\dfrac{JE}{JK}\implies\dfrac{(3x+21)-12}{12}=\dfrac{56}{70}[/tex]

Now we can solve for [tex]x[/tex]:

[tex]\dfrac{3x+9}{12}=\dfrac45[/tex]

[tex]3x+9=12\cdot\dfrac45=\dfrac{48}5[/tex]

[tex]3x=\dfrac{48}5-9=\dfrac35[/tex]

[tex]x=\dfrac{\frac35}3=\dfrac15[/tex]

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