Try this beautiful Problem on Algebra from the Linear equation from AMC 10 A, 2015.

**Linear Equation Problem **– AMC-10A, 2015- Problem 16

If $y+4=(x-2)^{2}, x+4=(y-2)^{2}$, and $x \neq y$, what is the value of $x^{2}+y^{2} ?$

,

- $11$
- $12$
- $15$
- $14$
- $6$

**Key Concepts**

Algebra

Equation

## Suggested Book | Source | Answer

#### Suggested Reading

Pre College Mathematics

#### Source of the problem

AMC-10A, 2015 Problem-16

#### Check the answer here, but try the problem first

$15$

## Try with Hints

#### First Hint

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 \(x+y\)?

#### Second Hint

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\( x \neq y\)]

\(\Rightarrow (x+y)=3\)

#### Third Hint

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

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=y-8Ru_qKDxk