Question #f23eb

1 Answer
Manikandan S.
Aug 15, 2017

Solution

Explanation:

From equation 1
#3x-2y=1#
#2y = 3x-1#

Equation 2 #=>#
#(x-2)^2+(2y+3)^2=26#
#x^2-4x+4+(3x+2)^2=26#
#x^2-4x+4+9x^2+12x+4=26#
#10x^2+8x-18 = 0#
#5x^2+4x-9 = 0#
#5x^2-5x+9x-9 = 0#
#5x(x-1)+9(x-1) = 0#
#(x-1)(5x+9) = 0#
#x = 1,-9/5#
#y = 1,-16/5#
Hence the solution is #(1,1)#,#(-9/5,-16/5)#

You can verify the result by substituting in the equations. For e.g.

Let us take the second set of solution and put them on equation 2
#(x-2)^2+(2y+3)^2=(-9/5-2)^2+(2*(-16/5)+3)^2#
#= ((-9-10)^2+(-32+15)^2)/5^2#
#=((-19^2)+(-17)^2)/25#
# = (361+289)/25#
#=650/25#
#=26=R.H.S#

Related questions
Impact of this question
740 views around the world
You can reuse this answer
Creative Commons License