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Algebra AMC 10 Math Olympiad USA Math Olympiad

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

,

  • $11$
  • $12$
  • $15$
  • $14$
  • $6$

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

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