Question

Line segment DB and line segment AE are medians. If BC=6y+10, AB=22+3y, CE=6x+12, ED=2x+60, then find the value of x and y, and the length of the segments.

Answer 1

Answer:

$x=45.5,\ y=\dfrac{280}{3}.$

Step-by-step explanation:

Line segment DB and line segment AE are medians, then

• $AD=DC=\dfrac{1}{2}AC;$
• $BE=CE=\dfrac{1}{2}BC.$.

This gives you that $6x+12=\dfrac{6y+10}{2}.$

By the triangle midline theorem,

$DE=\dfrac{1}{2}BC,$

then

$2x+60=\dfrac{22+3y}{2}.$

Solve the system of two equations:

$\left\{\begin{array}{l}6x+12=\dfrac{6y+10}{2}\\ \\2x+60=\dfrac{22+3y}{2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}12x+24=6y+10\\ \\4x+120=22+3y\end{array}\right.\Rightarrow \left\{\begin{array}{l}6x-3y+7=0\\ \\4x-3y+98=0\end{array}\right.$

Subtract the second equation from the first one:

$6x-3y+7-(4x-3y+98)=0,\\ \\2x=98-7,\\ \\2x=91,\\ \\x=45.5.$

Then

$4\cdot 45.5-3y+98=0,\\ \\3y=280,\\ \\y=\dfrac{280}{3}.$