Rectangles

If rectangle ABCD, diagonals AC and BD intersect at E. If AE=(2x-6y+37), EC=(4x-2y+5), and BD=(x+5y+7),
find the length of BD.

\(\begin{array}{|rcll|} \hline \mathbf{AE} &=& \mathbf{\dfrac{BD}{2}} \\\\ 2x-6y+37 &=& \dfrac{x+5y+7}{2} \\\\ 2(2x-6y+37) &=& x+5y+7 \\ 4x-12y+74 &=& x+5y+7 \\ \mathbf{3x-17y} &=& \mathbf{-67} \\ 3x &=& -67+17y \\ \mathbf{ x } &=& \mathbf{\dfrac{-67+17y}{3}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{EC} &=& \mathbf{\dfrac{BD}{2}} \\\\ 4x-2y+5 &=& \dfrac{x+5y+7}{2} \\\\ 2(4x-2y+5) &=& x+5y+7 \\ 8x-4y+10 &=& x+5y+7 \\ \mathbf{7x-9y} &=& \mathbf{-3} \quad | \quad \mathbf{ x=\dfrac{-67+17y}{3}} \\ 7\left(\dfrac{-67+17y}{3}\right)-9y &=& -3 \quad | \quad \times 3 \\ 7( -67+17y)-27y &=& -9 \\ -469+119y-27y &=& -9 \\ 92y &=& 469-9 \\ 92y &=& 460 \quad |\quad : 92 \\ \mathbf{ y } &=& \mathbf{5} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{ x } &=& \mathbf{\dfrac{-67+17y}{3}} \quad | \quad \mathbf{ y =5} \\ x &=& \dfrac{-67+17*5}{3} \\ x &=& \dfrac{18}{3} \\ \mathbf{ x } &=& \mathbf{6} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{BD} &=& x+5y+7 \quad | \quad x=6,\ y = 5 \\ \text{BD} &=& 6+5*5+7 \\ \mathbf{ \text{BD} } &=& \mathbf{38} \\ \hline \end{array}\)