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$$ (3x + 2)^2 + (3y+4)^2 = 0 $$ What is $ y+x $?

The answer is $-2$. Id like to know how. Thank you.

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  • $\begingroup$ For future reference, algebra is fundamentally different from linear algebra. $\endgroup$ Commented Sep 12, 2015 at 20:23

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In order for the two squares to be $0$. Both expressions $(3x+2)^2$ and $(3y+4)^2$ have to be $0$.

$$3x+2=0$$ $$3y+4=0$$

Add Both equations $3(x+y)+6=0$. I leave the rest to you.

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  • $\begingroup$ They can only be 0 and not compensating each other (-a and +a) because they are squares and thus $\ge0$ $\endgroup$
    – null
    Commented Sep 12, 2015 at 20:45

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