Linear Equation Problem | AMC 10A, 2015 | Problem No.16

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 \(x^2+y^2=5(x+y)=5 \times 3=15\)

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 \(x^2+y^2=5(x+y)=5 \times 3=15\)