Answer:
[tex]x=45.5,\ y=\dfrac{280}{3}.[/tex]
Step-by-step explanation:
Line segment DB and line segment AE are medians, then
- [tex]AD=DC=\dfrac{1}{2}AC;[/tex]
- [tex]BE=CE=\dfrac{1}{2}BC.[/tex].
This gives you that [tex]6x+12=\dfrac{6y+10}{2}.[/tex]
By the triangle midline theorem,
[tex]DE=\dfrac{1}{2}BC,[/tex]
then
[tex]2x+60=\dfrac{22+3y}{2}.[/tex]
Solve the system of two equations:
[tex]\left\{\begin{array}{l}6x+12=\dfrac{6y+10}{2}\\ \\2x+60=\dfrac{22+3y}{2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}12x+24=6y+10\\ \\4x+120=22+3y\end{array}\right.\Rightarrow \left\{\begin{array}{l}6x-3y+7=0\\ \\4x-3y+98=0\end{array}\right.[/tex]
Subtract the second equation from the first one:
[tex]6x-3y+7-(4x-3y+98)=0,\\ \\2x=98-7,\\ \\2x=91,\\ \\x=45.5.[/tex]
Then
[tex]4\cdot 45.5-3y+98=0,\\ \\3y=280,\\ \\y=\dfrac{280}{3}.[/tex]
