Given the triangle ABC, you can assume that DE is a Midsegment that connects the midpoints of two sides of the triangles.
According to the Triangle Midsegment Theorem, the Midsegment that is parallel to a side of the triangle has half the length of that side.
You can identify that DE and BC are parallel, then:
[tex]DE=\frac{BC}{2}[/tex]
Knowing that:
[tex]\begin{gathered} DE=3x+2 \\ BC=10x-4 \end{gathered}[/tex]
You can substitute the corresponding expressions into the first equation and solve for "x":
[tex]3x+2=\frac{10x-4}{2}[/tex][tex]3x+2=5x-2[/tex][tex]\begin{gathered} 3x-5x=-2-2 \\ \\ x=\frac{-4}{-2} \\ \\ x=2 \end{gathered}[/tex]
Hence, the answer is:
[tex]x=2[/tex]
