How do you solve this equation?

#x-2y=3#
#x^2+2y^2=27#

This is a simultaneous equation

1 Answer
maganbhai P.
Apr 28, 2018

#(i) x=-3+6sqrt2 and y=-3+3sqrt2#

#(ii) x=-3-6sqrt2 and y=-3-3sqrt2#

Explanation:

Here,

#color(red)(x-2y=3...to(I) and#

#color(red)(x^2+2y^2=27...to(II)#

Now, #x-2y=3=>color(red)(x=2y+3...to(III)#

From , #(II)#

#(2y+3)^2+2y^2=27#

#=>4y^2+12y+9+2y^2=27#

#=>2y^2+12y+9-27=0#

#=>2y^2+12y-18=0#

#=>y^2+6y-9=0#

Comparing with #ax^2+bx+c=0#

#a=1,b=6,c=-9#

#:.triangle=b^2-4ac=36-4(1)(-9)=36+36#

#sqrttriangle=sqrt(36xx2)=6sqrt2#

#:.y=(-b+-sqrt(triangle))/(2a)=(-6+-6sqrt2)/2=-3+-3sqrt2#

From #(III), #

#(i)y=-3+3sqrt2=>x=2(-3+3sqrt2)+3=-3+6sqrt2#

#=>color(blue)( x=-3+6sqrt2 and y=-3+3sqrt2#

#(ii)y=-3-3sqrt2=>x=2(-3-3sqrt2)+3=-3-6sqrt2#

#=>color(blue)( x=-3-6sqrt2 and y=-3-3sqrt2#

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