(i) x2 + xy – 3x – y + 2 = 0
(ii) x2 – y2 – 2x + 2y = 0
Solution:
(i) x2 + xy – 3x – y + 2 = 0
We will first substitute the value of x by x + 1 and y by y + 1. Then, te above-given ewuation becomes:
$ {{\left( x\text{ }+\text{ }1 \right)}^{2}}~+\text{ }\left( x\text{ }+\text{ }1 \right)\text{ }\left( y\text{ }+\text{ }1 \right)-3\left( x\text{ }+\text{ }1 \right)-\left( y\text{ }+\text{ }1 \right)\text{ }+\text{ }2\text{ }=\text{ }0 $
$ {{x}^{2}}~+\text{ }1\text{ }+\text{ }2x\text{ }+\text{ }xy\text{ }+\text{ }x\text{ }+\text{ }y\text{ }+\text{ }1-3x-3-y-1\text{ }+\text{ }2\text{ }=\text{ }0 $
Simplifying further we get,
x2 + xy = 0
Therefore, the transformed equation is x2 + xy = 0.
(ii) x2 – y2 – 2x + 2y = 0
We will first replace the value of x by x + 1 and y by y + 1
Then,
$ {{\left( x\text{ }+\text{ }1 \right)}^{2}}-{{\left( y\text{ }+\text{ }1 \right)}^{2}}-2\left( x\text{ }+\text{ }1 \right)\text{ }+\text{ }2\left( y\text{ }+\text{ }1 \right)\text{ }=\text{ }0 $
$ {{x}^{2}}~+\text{ }1\text{ }+\text{ }2x-{{y}^{2}}-1-2y-2x-2\text{ }+\text{ }2y\text{ }+\text{ }2\text{ }=\text{ }0 $
Simplifying further we get,
x2 – y2 = 0
Therefore, the transformed equation is x2 – y2 = 0.