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

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

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