From equation 1
3x-2y=13x−2y=1
2y = 3x-12y=3x−1
Equation 2 =>⇒
(x-2)^2+(2y+3)^2=26(x−2)2+(2y+3)2=26
x^2-4x+4+(3x+2)^2=26x2−4x+4+(3x+2)2=26
x^2-4x+4+9x^2+12x+4=26x2−4x+4+9x2+12x+4=26
10x^2+8x-18 = 010x2+8x−18=0
5x^2+4x-9 = 05x2+4x−9=0
5x^2-5x+9x-9 = 05x2−5x+9x−9=0
5x(x-1)+9(x-1) = 05x(x−1)+9(x−1)=0
(x-1)(5x+9) = 0(x−1)(5x+9)=0
x = 1,-9/5x=1,−95
y = 1,-16/5y=1,−165
Hence the solution is (1,1)(1,1),(-9/5,-16/5)(−95,−165)
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(x−2)2+(2y+3)2=(−95−2)2+(2⋅(−165)+3)2
= ((-9-10)^2+(-32+15)^2)/5^2=(−9−10)2+(−32+15)252
=((-19^2)+(-17)^2)/25=(−192)+(−17)225
= (361+289)/25=361+28925
=650/25=65025
=26=R.H.S=26=R.H.S
