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Algebra Difficulty 3.7 AMC 10/12 Find the answer Canada

If $x^2=8x+y$ and $y^2=x+8y$ with $x\ne y$ , then the value of $x^2+y^2$ is

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Solution

Since $x^2 = 8x+y$ and $y^2=x+8y$ , then $x^2-y^2 = (8x+y)-(x+8y) = 7x - 7y$ .

Factoring both sides, we obtain $(x+y)(x-y) = 7(x-y)$ .

Since $x \neq y$ , then $x-y \neq 0$ , so we can divide both sides by $x-y$ to obtain $x+y = 7$ .

Since $x^2 = 8x+y$ and $y^2=x+8y$ , then $x^2+y^2 = (8x+y)+(x+8y)=9x+9y = 9(x+y)=9 \cdot 7 = 63$

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.