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