High School Math Question (Version 2)

$x^2 - y^2 + x^3 - y^3 = 494$

$(x-y)(x+y) + (x-y)(x^2 + xy + y^2) = 494$

$(x-y)(x^2 + xy + y^2 + x + y) = 494$

$(x-y) \times 494 = 494$

$(x-y) = 1$

Now, $x^2 - y^2 = 25$

$(x-y)(x+y) = 25$

$(x+y) = \boxed{25}$

$$ Rewrite: $$ $\begin{aligned} x^2-y^2=25&\quad\Rightarrow\quad(x+y)(x-y)=25\\ x^2+x+xy+y+y^2=494&\quad\Rightarrow\quad(x+y)+(x^2+xy+y^2)=494\\ x^3-y^3=469&\quad\Rightarrow\quad(x-y)(x^2+xy+y^2)=469 \end{aligned}$ $$ Let $\;a=x+y$ , $\;b=x-y$ , and $\;c=x^2+xy+y^2$ , then $$ $\begin{aligned} ab=25&\quad\quad\quad(1)\\ a+c=494&\quad\quad\quad(2)\\ bc=469&\quad\quad\quad(3) \end{aligned}$ $$ from $(1)$ and $(3)$ , we obtain $$ $\begin{aligned} \frac{bc}{ab}&=\frac{469}{25}\\ \frac{c}{a}&=\frac{469}{25}\\ c&=\frac{469}{25}a \end{aligned}$ $$ Pluging in the last part to $(2)$ , yield $$ $\begin{aligned} a+\frac{469}{25}a&=494\\ \frac{494}{25}a&=494\\ a&=25. \end{aligned}$ $$ Thus, $a=x+y=\boxed{25}$ .

x^2 + y^2 + xy + (x + y) = 494---------(iii) Now, x^2 + y^2 + xy = (x^3 - y^3) / (x - y) = 469 / (x - y)---(i) Now , (x - y)(x + y) = 25 , (x - y) = 25 / (x + y)---(ii) Substituting (ii) in (i) , 469 / ( 25 / (x + y) ) = 469(x+y)/25 Let , (x + y) = z Putting all values in (iii) 469z/25 + z = 494 469z + 25z = 12350 494z = 12350 z = 12350 / 494 z = 25 But , z = (x + y) Ans: (x + y) = 25

its damn easy... by common sense, we know that 13^2 - 12^2 = 25. (remember Pythagoras triplets) so x=13 and y=13. so x+y= 25.

x^2+x+xy+y+y^2=494, (x-y)(x^2+x+xy+y+y^2)=494(x-y), (x-y)(x^2+xy+y^2)+(x+y)(x-y)=494(x-y), (x^3-y^3)+x^2-y^2=494(x-y), 469+25=494(x-y), 494=494(x-y), x-y=1. Again, x^2-y^2=25, (x+y)(x-y)=25, (x+y)*1=25, x+y=25.