Question 1008264


 In rectangle {{{ABCD}}}, diagonals {{{AC}}} and {{{BD}}} intersect at {{{E}}}.
 
If {{{AE= 3x+y}}}, {{{BE= 4x-2y}}}, and {{{CE= 20}}}, 
find {{{x}}} and {{{y}}}:

since you have a rectangle, diagonals {{{AC}}} and {{{BD}}} are same length

then{{{AE= CE}}} and  {{{BE= CE}}}

so, we have

{{{ 3x+y=20}}}....eq.1

{{{ 4x-2y=20}}}....eq.2
----------------------...both sides of eq.1 multiply by {{{2}}}



{{{ 6x+2y=40}}}....eq.1

{{{ 4x-2y=20}}}....eq.2
----------------------........add both eq.1 and eq.2

{{{ 6x+cross(2y)+4x-cross(2y)=40+20}}} 

{{{ 10x=60}}} 

{{{ x=60/10}}} 

{{{ highlight(x=6)}}} 

now find {{{y}}}

{{{ 3x+y=20}}}....eq.1

{{{ y=20-3x}}} 

{{{ y=20-3*6}}}

{{{ y=20-18}}}

{{{highlight( y=2)}}}