Try this beautiful Problem on Algebra from the Linear equation from AMC 10 A, 2015.
If $y+4=(x-2)^{2}, x+4=(y-2)^{2}$, and $x \neq y$, what is the value of $x^{2}+y^{2} ?$
,
Algebra
Equation
Pre College Mathematics
AMC-10A, 2015 Problem-16
$15$
Given that $y+4=(x-2)^{2}, x+4=(y-2)^{2}$ . we have to find out $x^{2}+y^{2} ?$. Now add two equations $x^{2}+y^{2}-4 x-4 y+8=x+y+8$
\(\Rightarrow x^{2}+y^{2}=5(x+y)\)
Can you find out the value \(x+y\)?
We can also subtract the two equations to yield the equation
$x^{2}-y^{2}-4 x+4 y=y-x$
\(\Rightarrow x^{2}-y^{2}=(x+y)(x-y)=3 x-3 y=3(x-y)\)
Therefore \((x+y)(x-y)=3 x-3 y=3(x-y)\)
\(\frac{(x+y)(x-y)}{(x-y)}=\frac{3(x-y)}{(x-y)}\) [ as\( x \neq y\)]
\(\Rightarrow (x+y)=3\)
Therefore \(x^2+y^2=5(x+y)=5 \times 3=15\)
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