Categories

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

Try this beautiful Problem based on Linear Equations, Algebra AMC 10A, 2015, Problem-16. You may use sequential hints to solve the problem.

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 $x^{2}+y^{2} ?$

,

• $11$
• $12$
• $15$
• $14$
• $6$

Algebra

Equation

## Suggested Book | Source | Answer

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